Question

Proof that a cube plus + b cube + eight c cube =6abc​

Answer 1

Step-by-step explanation:

a+b+2c = 0

We know , a³ + b³ + 8c³ - 3(a)(b)(2c)

= a³ + b³ + 8c³ - 6abc

= (a+b+2c) { a²+b²+4c² - (ab+2bc+ca) }

= 0 × { -2(ab+2bc+ca) - (ab+2bc+ca) }

= 0 × { -3(ab+2bc+ca) }

= 0

Now , a³ + b³ + 8c³ - 6abc = 0

So , a³ + b³ + 8c³ = 6abc

Hence proof