Question

3√3a3-b3-5√5c3-3√15abc

Answer 1

Hi ,

*****************************************

We know the algebraic identity :

x³ + y³ + z³ - 3xyz

= (x+y+z)(x²+y²+z²-xy-yz-zx )

*******************************************
Here ,

x = √3a

y = -b ,

z = -√5c;

Therefore,

3√3a³-b³-5√5c³-3√15abc

= (√3a )³ + ( - b )³ + ( -√5c )-3×√3a×(-b)×(-√5c)

=(√3a-b-√5c)[(√3a)²+(-b)²+(-√5c)²-√3a(-b)-(-b)(-√5c)-(-√5c)(√3a )]
= (√3a-b-√5c)[3a²+b²+5c²+√3ab-√5bc+√15ac]

I hope this helps you.

: )