Question

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

Answer 1

Step-by-step explanation:

Given :-

a + b + 2c = 0

To find :-

Prove that a³ + b³ + 8c³ = 6abc

Solution :-

Given that :

a + b + 2c = 0

=> a + b = -2c --------(1)

On cubing both sides then

=> (a + b)³ = (-2c)³

=> a³ + b³ + 3ab (a+b) = -2³c³

=> a³ + b³ + 3ab (a+b) = -8c³

=> a³ + b³ + 3ab (- 2c ) = -8c³

(from (1))

=> a³ + b³ - 6abc = -8 c³

=> a³ + b³ + 8c³ = 6abc

Alternative Method :-

We know that

If a + b + c = 0 then a³ + b³ + c³ = 3abc

We have a + b + 2c = 0

Where a = a , b = b and c = 2c

Now,

If a + b + 2c = 0 then

a³ + b³ + (2c)³ = 3(a)(b)(2c)

=>a³ + b³ + 8c³ = 6abc

Hence, Proved.

Answer:-

If a + b + 2c = 0 then a³ + b³ + 8c³ = 6abc

Used formulae :-

• (a+b)³ = a³ + b³ + 3ab (a+b)

• If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

Answer:

1 ,2 ka 4

4 2 ka 1

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