Question

factorise the following

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Answer 1

Answer:

this is the answer

Step-by-step explanation:

follow me

Answer 2

$\huge\mathbb{\underline{QUESTION:-}}$

$\rm (p{}^{2}-2pq+q{}^{2})-r{}^{2}$

$\huge\mathbb{\underline{SOLUTION:-}}$

Star with the simplification

. Identity will be used :-

$\sf 1) (a-b){}^{2}=a{}^{2}-2ab+b{}^{2}\\ \sf 2) a{}^{2}-b{}^{2}=(a+b)(a-b)$

$\rm (p{}^{2}-2pq+q{}^{2})-r{}^{2}$

$\rm\implies (p{}^{2}-2pq+q{}^{2}):-(p-q){}^{2}$

$\rm Now \: apply \:this\: identity$

$\rm\implies (p-q){}^{2}-r{}^{2}$

$\rm\implies It \:is\:in\:the\:form\:of :-$

$\rm a{}^{2}-b{}^{2}:-(a+b)(a-b)$

$\rm where\:a=(p-q) and \:b= r$

$\rm Now\:-\\ \rm (p-q){}^{2}-r{}^{2}\\ \rm (p-q+r)(p-q-r)$

So, the answer is (p-q+r)(p-q-r)

$\large\boxed{\red{\sf{(p-q+r)(p-q-r)}}}$