If $a = 1$ and $b = - 1$ , find the value of the following :
$(a^{2} + b^{2}) (- a^{2} + b^{2})$

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Solution

Put $a = 1, b = - 1$ in $(a^{2} + b^{2}) (- a^{2} + b^{2})$
$= {1^{2} + (- 1)^{2}} {- (1)^{2} + (- 1)^{2}} [∵ (- 1)^{2} = 1]$
$= (1 + 1) (- 1 + 1)$
$= 2 \times 0 = 0$
Hence, $(a^{2} + b^{2}) (- a^{2} + b^{2}) = 0$

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