Question

if a+b+c=0 tgen find the value of a+³b³+c³​

Answer 1

a3+b3+c3=3abc

Hope it helped you mate

Answer 2

As we know,
(a+b+c)³ = (a+b+c)² (a+b+c)

→(a+b+c)³ = a³ + b³ + c³ + a²(b+c) + b²(a+c) + c²(a+b) + 2(ab+bc+ac)(a+b+c)

= a³ + b³ + c³ + 3a²(b+c) + 3b²(a+c) + 3c²(a+b) + 6abc

So, (a+b+c)³ = a³ + b³ + c3 + 3(a+b)(b+c)(a+c)

From the last step:
→ a³ + b³ + c³ = (a+b+c)³ − [3a²(b+c) + 3b²(a+c) + 3c²(a+b) + 6abc]

So, a³ + b³ + c³ − 3abc
=(a+b+c)³ − [3a²(b+c) + 3b²(a+c) + 3c²(a+b) + 9abc]

Split 9abc among the three terms and now collect ab, bc & ac terms:

= (a+b+c)³ − [3ab(a+b+c) + 3bc(a+b+c) + 3ac(a+b+c)]

Take (a+b+c) as common outside,
=(a+b+c)[a²+b²+c²+2ab+2bc+2ac−3ab−3bc−3ac]

Thus we get:
a³ + b³ + c³ − 3abc=(a+b+c)[a² + b² + c² −ab−bc−ac]

which may further be rewritten as:
a³ + b³ + c³ − 3abc=(a+b+c)(1/2)[(a−b)²+(b−c)²+(a−c)²]

as (a−b)²=a²+b²−2ab

If a+b+c=0.
Clearly the RHS =0
So, a³ + b³ + c³ = 3abc

Thankyou!!!