Show that: ${(9 p - 5 q)}^{2} + 180 pq = {(9 p + 5 q)}^{2}$

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Solution

To prove : ${(9 p - 5 q)}^{2} + 180 pq = {(9 p + 5 q)}^{2}$
$L . H . S$ $: {(9 p - 5 q)}^{2} + 180 pq$
$= {(9 p)}^{2} + {(5 q)}^{2} - 2 \times 9 p \times 5 q + 180 p q$ [ Formula used : ${(a - b)}^{2} = a^{2} + b^{2} - 2 ab$ ]
$= {(9 p)}^{2} + {(5 q)}^{2} - 90 p q + 180 p q$
$= {(9 p)}^{2} + {(5 q)}^{2} + 90 p q$
$= {(9 p)}^{2} + {(5 q)}^{2} + 2 \times 9 p \times 5 q$
$= {(9 p + 5 q)}^{2}$ [ Formula used : ${(a + b)}^{2} = a^{2} + b^{2} + 2 ab$ ]
From above it is clear that $L . H . S$ $=$ $R . H . S$
Thus, ${(9 p - 5 q)}^{2} + 180 pq = {(9 p + 5 q)}^{2}$
Hence Proved.

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