factorise (x^2-2xy+y)^2-z^2
brunoconti:
is it y or y^2?. i Will consider y^2
Answers
Answered by
30
We have (x2 - 2xy + y2) = (x - y)2
∴ (x2 - 2xy + y2) - z2 = (x - y)2 - (z)2
= [(x - y) + z] [(x - y) - z]
[Using a2 - b2 = (a + b) (a - b)]
= (x - y + z)(x - y - z)
∴ (x2 - 2xy + y2) - z2 = (x - y + z) (x - y -z)
∴ (x2 - 2xy + y2) - z2 = (x - y)2 - (z)2
= [(x - y) + z] [(x - y) - z]
[Using a2 - b2 = (a + b) (a - b)]
= (x - y + z)(x - y - z)
∴ (x2 - 2xy + y2) - z2 = (x - y + z) (x - y -z)
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Birlianist
Answered by
2
Answer:
(x-y+z)(x-y-z)
Step-by-step explanation:
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