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# 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)$\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$

Last updated date: 13th Jul 2024
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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.