Question

question no. 6 please

Answer 1

Here is your solution,

2√2a³ + 3√3b³ + c³ + 3√6abc

I t is in the form of a³+b³+c³+3abc = (a+b+c) (a²+b²+c²-ab-bc-ca)

Here, a = √2a b = √3b c = c

= (√2a+√3b+c) [(√2a)² + (√3b)² + (c)²-(√2a.√3√b)-(√3b.c)-(c.√2a)

=(√2a+√3b+c)(2a∧2+3b∧2+c∧2-√6ab-√3bc-√2ac)