Show that:
(i) ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$
(ii)${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$
(iii)${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$
(iv)${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$
(v)$\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$
Answer
381.3k+ views
Hint: The given statements resemble the following algebraic identities.
${\left( {a + b} \right)^2} - 4ab = {\left( {a - b} \right)^2}$, ${\left( {a - b} \right)^2} + 4ab = {\left( {a + b} \right)^2}$, ${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab$
Use these to prove the given statements.
Let us prove the statements one by one using the identities.
(i) ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$
We know that, ${\left( {a + b} \right)^2} - 4ab = {\left( {a - b} \right)^2}$ …(1)
Let us prove this identity first.
LHS=${\left( {a + b} \right)^2} - 4ab$
Expand the brackets using the identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$
= {a^2} + {b^2} + 2ab - 4ab \\
= {a^2} + {b^2} - 2ab \\
$
We know that${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Using this in the previous step,
$ = {\left( {a - b} \right)^2}$
=RHS
Hence, the identity is proved. Now let us use this to prove the given statement.
${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$ is the given statement to be proved.
LHS$ = {\left( {3x + 7} \right)^2} - 84x$
Using identity (1) here, we find that $a = 3x,b = 7$
Substituting it in the LHS of (1) and using the identity, we get
\[
LHS = {\left( {3x + 7} \right)^2} - 4\left( {3x} \right)\left( 7 \right) \\
= {\left( {3x + 7} \right)^2} - 84x \\
= {\left( {3x - 7} \right)^2} \\
= RHS \\
\]
Hence, ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$is proved.
(ii)${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$
We know that, ${\left( {a - b} \right)^2} + 4ab = {\left( {a + b} \right)^2}$ …(2)
Let us prove this identity first.
LHS=${\left( {a - b} \right)^2} + 4ab$
Expand the brackets using the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$
= {a^2} + {b^2} - 2ab + 4ab \\
= {a^2} + {b^2} + 2ab \\
$
We know that${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Using this in the previous step,
$ = {\left( {a + b} \right)^2}$
=RHS
Hence, the identity is proved. Now let us use this to prove the given statement.
${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$ is the given statement to be proved.
LHS$ = {\left( {9p - 5q} \right)^2} + 180pq$
Using identity (2) here, we find that $a = 9p,b = 5q$
Substituting it in the LHS of (2) and using the identity, we get
\[
LHS = {\left( {9p - 5q} \right)^2} + 4\left( {9p} \right)\left( {5q} \right) \\
= {\left( {9p - 5q} \right)^2} + 180pq \\
= {\left( {9p + 5q} \right)^2} \\
= RHS \\
\]
Hence, ${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$is proved.
(iii)${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$
We know that, ${\left( {a - b} \right)^2} + 4ab = {\left( {a + b} \right)^2}$
We have already proved this identity.
Using this identity here, we find that $a = \dfrac{4}{3}m,b = \dfrac{3}{4}n$
Substituting it in the LHS of (2) and using the identity, we get
\[
LHS = {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 4\left( {\dfrac{4}{3}m} \right)\left( {\dfrac{3}{4}n} \right) \\
= {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 4mn \\
= {\left( {\dfrac{4}{3}m + \dfrac{3}{4}n} \right)^2} \\
= \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2} + 2\left( {\dfrac{4}{3}m} \right)\left( {\dfrac{3}{4}n} \right) \\
= \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2} + 2mn \\
\]
Now we have proved ${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 4mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2} + 2mn$
Rearranging it we get,
${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$
Hence, ${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$is proved.
(iv)${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$
From identity (2), by rearranging it we get,
${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab$ …(3)
Using identity (3) here, we find that $a = 4pq,b = 3q$
Substituting it in the LHS of (3) and using the identity, we get
$
{\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = {\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} \\
= 4\left( {4pq} \right)\left( {3q} \right) \\
= 48p{q^2} \\
= RHS \\
$
Hence,${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$is proved.
(v)$\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$
We know that, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ …(4)
Using this identity, we need to open the brackets for the LHS of the given statement.
$
LHS = \left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) \\
= {a^2} - {b^2} + {b^2} - {c^2} + {c^2} - {a^2} \\
$
After cancelling the terms, we get
$LHS = 0 = RHS$
Hence,$\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$is proved.
Note: These statements can also be proved without the use of the identities (1), (2) and (3) by just expanding the brackets and by simplifying it. But it would have been a little lengthier process than by using the identities.
${\left( {a + b} \right)^2} - 4ab = {\left( {a - b} \right)^2}$, ${\left( {a - b} \right)^2} + 4ab = {\left( {a + b} \right)^2}$, ${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab$
Use these to prove the given statements.
Let us prove the statements one by one using the identities.
(i) ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$
We know that, ${\left( {a + b} \right)^2} - 4ab = {\left( {a - b} \right)^2}$ …(1)
Let us prove this identity first.
LHS=${\left( {a + b} \right)^2} - 4ab$
Expand the brackets using the identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$
= {a^2} + {b^2} + 2ab - 4ab \\
= {a^2} + {b^2} - 2ab \\
$
We know that${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Using this in the previous step,
$ = {\left( {a - b} \right)^2}$
=RHS
Hence, the identity is proved. Now let us use this to prove the given statement.
${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$ is the given statement to be proved.
LHS$ = {\left( {3x + 7} \right)^2} - 84x$
Using identity (1) here, we find that $a = 3x,b = 7$
Substituting it in the LHS of (1) and using the identity, we get
\[
LHS = {\left( {3x + 7} \right)^2} - 4\left( {3x} \right)\left( 7 \right) \\
= {\left( {3x + 7} \right)^2} - 84x \\
= {\left( {3x - 7} \right)^2} \\
= RHS \\
\]
Hence, ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$is proved.
(ii)${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$
We know that, ${\left( {a - b} \right)^2} + 4ab = {\left( {a + b} \right)^2}$ …(2)
Let us prove this identity first.
LHS=${\left( {a - b} \right)^2} + 4ab$
Expand the brackets using the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$
= {a^2} + {b^2} - 2ab + 4ab \\
= {a^2} + {b^2} + 2ab \\
$
We know that${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Using this in the previous step,
$ = {\left( {a + b} \right)^2}$
=RHS
Hence, the identity is proved. Now let us use this to prove the given statement.
${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$ is the given statement to be proved.
LHS$ = {\left( {9p - 5q} \right)^2} + 180pq$
Using identity (2) here, we find that $a = 9p,b = 5q$
Substituting it in the LHS of (2) and using the identity, we get
\[
LHS = {\left( {9p - 5q} \right)^2} + 4\left( {9p} \right)\left( {5q} \right) \\
= {\left( {9p - 5q} \right)^2} + 180pq \\
= {\left( {9p + 5q} \right)^2} \\
= RHS \\
\]
Hence, ${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$is proved.
(iii)${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$
We know that, ${\left( {a - b} \right)^2} + 4ab = {\left( {a + b} \right)^2}$
We have already proved this identity.
Using this identity here, we find that $a = \dfrac{4}{3}m,b = \dfrac{3}{4}n$
Substituting it in the LHS of (2) and using the identity, we get
\[
LHS = {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 4\left( {\dfrac{4}{3}m} \right)\left( {\dfrac{3}{4}n} \right) \\
= {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 4mn \\
= {\left( {\dfrac{4}{3}m + \dfrac{3}{4}n} \right)^2} \\
= \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2} + 2\left( {\dfrac{4}{3}m} \right)\left( {\dfrac{3}{4}n} \right) \\
= \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2} + 2mn \\
\]
Now we have proved ${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 4mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2} + 2mn$
Rearranging it we get,
${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$
Hence, ${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$is proved.
(iv)${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$
From identity (2), by rearranging it we get,
${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab$ …(3)
Using identity (3) here, we find that $a = 4pq,b = 3q$
Substituting it in the LHS of (3) and using the identity, we get
$
{\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = {\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} \\
= 4\left( {4pq} \right)\left( {3q} \right) \\
= 48p{q^2} \\
= RHS \\
$
Hence,${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$is proved.
(v)$\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$
We know that, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ …(4)
Using this identity, we need to open the brackets for the LHS of the given statement.
$
LHS = \left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) \\
= {a^2} - {b^2} + {b^2} - {c^2} + {c^2} - {a^2} \\
$
After cancelling the terms, we get
$LHS = 0 = RHS$
Hence,$\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$is proved.
Note: These statements can also be proved without the use of the identities (1), (2) and (3) by just expanding the brackets and by simplifying it. But it would have been a little lengthier process than by using the identities.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Scroll valve is present in a Respiratory system of class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

is known as the Land of the Rising Sun A Japan B Norway class 8 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
