Question

factorise (p²+q²-r²)² - 4p²q²​

Answer 1

Answer:

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\left(p^{2}+q^{2}-r^{2}\right)^{2}-4p^{2}q^{2}

(p2+q2−r2)2−4p2q2

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(p^{2}+q^{2}-r^{2})(p^{2}+q^{2}-r^{2})-4p^{2}q^{2}

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Step-by-step explanation:

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Answer 2

Answer:

$\mathrm{Factor}\:\left(p^2+q^2-r^2\right)^2-4p^2q^2:\quad \left(p^2+q^2-r^2+2pq\right)\left(p^2+q^2-r^2-2pq\right)$

Step-by-step explanation:

$\left(p^2+q^2-r^2\right)^2-4p^2q^2$

$=\left(p^2+q^2-r^2\right)^2-\left(2pq\right)^2$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\left(p^2+q^2-r^2\right)^2-\left(2pq\right)^2=\left(\left(p^2+q^2-r^2\right)+2pq\right)\left(\left(p^2+q^2-r^2\right)-2pq\right)$

$=\left(\left(p^2+q^2-r^2\right)+2pq\right)\left(\left(p^2+q^2-r^2\right)-2pq\right)$

$=\left(p^2+2pq+q^2-r^2\right)\left(p^2-2pq+q^2-r^2\right)$

HOPE IT HELPS YOU

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