Question

(x6-y6) by (x-y)
${x}^6{y {}^6{} } by x - y$

Answer 1

$<u><b><huge>Hello<u><b><huge>$

ANSWER:-

Using Some Property;

1) [x³-y³]

(x-y)(x²+x-y+y²)

2)[x³+y³]

(x+y)(x²-xy+y²)

Thanks!!!?

Answer 2

First of all, we should factorise the first polynomial in order to make it easier

So,
${x}^{6} - {y}^{6} \\ \\ = > ( {x}^{3} ) {}^{2} - (y {}^{3} ) {}^{2}$
now using, a² - b² = (a + b)(a - b)

we get,
$( {x}^{3} ) {}^{2} - ( {y}^{3} ){}^{2} \\ \\ = > ( {x}^{3} - {y}^{3} )( {x}^{3} + {y}^{3} )$
now using the identity,

a³ + b³ = (a + b)(a² + b² - ab)

and

a³ - b³ = (a - b)(a² + b² + ab)

we get,

$(x - y)( {x}^{2} + {y}^{2} + xy)(x + y)( {x}^{2} + {y}^{2} - xy)$
we have to divide this by (x - y) right

So when dividing we get,

$\frac{(x - y)( {x}^{2} + {y}^{2} + xy)(x + y)( {x}^{2} + {y}^{2} - xy) }{(x - y)}$

Cancellation of (x - y) in the denominator by the numerator we get,

$( {x}^{2} + {y}^{2} + xy)(x + y)( {x}^{2} + {y}^{2} - xy) \\ \\ or \\ \\ ( {x}^{2} + {y}^{2} + xy)( {x}^{3} + y {}^{3} )$
If you want to solve further,

$( {x}^{2} + {y}^{2} + xy)( {x}^{3} + y {}^{3} ) \\ \\ = > {x}^{5} + {x}^{3} {y}^{2} + {x}^{4} y + {y}^{3} {x}^{2} + {y}^{5} + {y}^{4} x$
hope it helps dear friend ☺️