nbsp;(x^2 nbsp;– 2xy + y^2) – z^2 = (x – y)^2 – z^2[∵ (a nbsp;– b)^2 nbsp;= a^2 – 2ab+b^2] = (x – y – z) (x – y + z) [∵ a^2 nbsp;– b^2 nbsp;= (a + ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Division of Polynomials

Factorise:𝒙�...

Question

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

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(x - y - z) (x - y + z)$
$(x^{2} - 2 x y + y^{2}) - z^{2}$
$= (x - y)^{2} - z^{2}$ $[∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= (x - y - z) (x - y + z)$
$[∵ a^{2} - b^{2} = (a + b) (a - b)]$

Suggest Corrections

Join BYJU'S Learning Program

Factorise: (x2−2xy+y2)−z2

The correct option is A (x−y−z)(x−y+z) (x2–2xy+y2)–z2 =(x–y)2–z2[∵(a–b)2=a2–2ab+b2] =(x–y–z)(x–y+z) [∵a2–b2=(a+b)(a–b)]

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

The correct option is A $(x - y - z) (x - y + z)$
$(x^{2} - 2 x y + y^{2}) - z^{2}$
$= (x - y)^{2} - z^{2}$ $[∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= (x - y - z) (x - y + z)$
$[∵ a^{2} - b^{2} = (a + b) (a - b)]$