Question

8 If a + b + c = 0, show that: a³+ b³ + c³ = 3abc.​

Answer 1

Step-by-step explanation:

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Given that the value of a+b+c=0

Putting the value in the equation we get

a^3+b^3+c^3-3abc=0(a^2+b^2+c^2-ab-bc-ca)

=0

Therefore,

a^3++b^3+c^3=3abc

HENCE PROVED

Answer 2

Step-by-step explanation:

Given:

a + b + c = 0

As we know that:

${a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

Putting a+b+c = 0

We will get

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (0)( { a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0$

Now Transferring -3abc to RHS we will get

${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

Hence Proved.

Hope it will be helpful.