1.

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

Answer»

The factorisation of (<klux>X</klux>+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} -<klux>B</klux>^{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)".


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