Question

x+y+z whole square - x-y-z whole square

Answer 1

The factorisation of $(x+y+z)^{2}-(x-y-z)^{2}$ is equal to "4x(y + z)".

Step-by-step explanation:

We have,

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

To find, the factorisation of $(x+y+z)^{2}-(x-y-z)^{2}$ = ?

∴ $(x+y+z)^{2}-(x-y-z)^{2}$

Using the algebraic identity,

$a^{2} -b^{2} =(a+b)(a-b)$

Here, a = x + y + z and b = x - y - z

= [(x + y + z)+(x - y - z)][(x + y + z)- (x - y - z)]

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

= (x + x)( y + z+ y + z)

= (2x)( 2y + 2z)

= (2)(2)(x)(y + z)

= 4x(y + z)

∴ The factorisation of $(x+y+z)^{2}-(x-y-z)^{2}$ = 4x(y + z)

Thus, the factorisation of $(x+y+z)^{2}-(x-y-z)^{2}$ is equal to "4x(y + z)".

Answer 2

Answer:

4X(Y+Z)

FINE? ✅

it took me a lot of time to solve!