Question

If a+b+2c = 0 prove that a3+b+8c3 = 6abc​

Answer 1

Question should be:

• If a+b+2c = 0 prove that a³+b³+8c³ = 6abc

GIVEN:

• a+b+2c = 0

TO PROVE:

• a³+b³+8c³ = 6abc

SOLUTION:

=> a+b+2c = 0

=> a+b = (-2c) .....(i)

Cubing both sides we get;

=> (a+b)³ = (-2c)³

=> a³+b³+3ab(a+b) = -8c³

Putting (a+b) = -2c from (i)

=> a³+b³+3ab(-2c) = -8c³

=> a³+b³-6abc = -8c³

=> a³+b³+8c³ = 6ab

PROVED:

NOTE:

Some important formulas:

(a+b)² = a²+b²+2ab

(a-b)² = a²+b²-2ab

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

a³+b³ = (a+b)(a²+b²-ab)

a³-b³ = (a-b)(a²+b²+ab)

(a+b)² = (a-b)²+4ab

(a-b)² = (a+b)²-4ab