Factorisation of z2-x2-2 x y-y2-A- x-y-zz-x-yB- x-y-zz-x-yC- x-y-zz-x-yD- x-y-zz-x-y

The correct option is D (x+y+z)(z x y) nbsp;Given: z^2 (x^2 + 2xy + y^2). Using the identity (a + b)^2=a^2 + 2ab + b^2 in the above equation we get, nbs ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation Using Algebraic Identities

Factorisation...

Question

Factorisation of $z^{2} - (x^{2} + 2 x y + y^{2})$ = .

(x + y - z) (z - x - y)

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(x + y + z) (z + x - y)

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(x + y + z) (z - x + y)

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(x + y + z) (z - x - y)

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Solution

The correct option is D $(x + y + z) (z - x - y)$

Given: $z^{2} - (x^{2} + 2 x y + y^{2}) .$

Using the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ in the above equation we get,
$z^{2} - (x^{2} + 2 x y + y^{2}) = z^{2} - (x + y)^{2}$ . . . (i)

Using the identity $a^{2} - b^{2} = (a + b) (a - b)$ in (i) we get,
$z^{2} - (x + y)^{2} = [z + (x + y)] [z - (x + y)]$
$= (x + y + z) (z - x - y)$

Suggest Corrections

Factorisation of z2−(x2+2xy+y2) = .

The correct option is D (x+y+z)(z−x−y) Given: z2−(x2+2xy+y2). Using the identity (a+b)2=a2+2ab+b2 in the above equation we get, z2−(x2+2xy+y2)=z2−(x+y)2 . . . (i) Using the identity a2−b2=(a+b)(a−b) in (i) we get, z2−(x+y)2=[z+(x+y)][z−(x+y)] =(x+y+z)(z−x−y)

Factorisation of $z^{2} - (x^{2} + 2 x y + y^{2})$ = .