Factorisation of nbsp- z-2 - -x-2 - 2xy - y-2- - nbsp-x-y-z-z-x-y- nbsp-x-y-z-z-x-y- nbsp-x-y-z-z-x-y- nbsp-x-y-z-z-x-y- nbsp-

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 VII

Mathematics

Factorisation of Algebraic Expressions Using 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,
$z^{2} - (x + y)^{2} = [z + (x + y)] [z - (x + y)]$
$= (x + y + z) (z - x - y)$

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Similar questions

Q. Factorisation of

z^{2} - (x^{2} + 2 x y + y^{2})

= .

Q. Using properties of determinants, prove that

ω ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z x^{2} & y^{2} & z^{2} y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x + y + z)

Q. If

x + y + z = 0

,find

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z x^{2} & y^{2} & z^{2} y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ ∣ =

Q. Prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{3} 1 & y & y^{3} 1 & z & z^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x + y + z) (x - y) (y - z) (z - x)

Q. If

x + y + z = 0

,then prove that

\frac{x y z}{(x + y) (y + z) (z + x)} = - 1

where (

x \neq - y, y \neq - z, z \neq - x

) ?

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