Question

factorise... plz solve both

Answer 1

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THE ANSWER IS HERE,

ix)

$= > \: 125 {(x - y)}^{3} + {(5y - 3z)}^{3} + {(3z - 5x)}^{3}$

$= > \: {5}^{3} {(x - y)}^{3} + {(5y - 3z)}^{3} +( {3z -5x) }^{3}$

$= > \: {(5x - 5y)}^{3} + ( {5y - 3z)}^{3} + {(3z - 5x)}^{3}$

This is in the form of

$= > \: {a}^{3} + {b}^{3} + {c}^{3}$
Let's find a+b+c.

=> (5x-5y)+(5y-3z)+(3z-5x)

=> 5x-5y+5y-3z+3z-5x

=> 0.

a+b+c=0.

So,
if a+b+c=0. Then

$= > \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

=> 3 (5x-5y)(5y-3z)(3z-5x)

x)

Given that,

$= > \: {(x - y)}^{3} + {(y - z)}^{3} + {(z - x)}^{3}$

This is in the form of

$= > \: {a}^{3} + {b}^{3} + {c}^{3}$

So, let's find the value of a+b+c.

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

=> x-y+y-z+z-x

=> 0.

a+b+c=0.

So, if a+b+c=0.Then,

$= > \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

=> (x-y)(y-z)(z-x).


:-)Hope it helps u.