Answer
Verified
480.3k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE