Question

a=-2,b=-3 and c=5 then find value of a^3+b^3+c^3-3abc

Answer 1

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Answer 2

Answer:

0

Step-by-step explanation:

a=-2,b=-3 and c=5,

So A+b+c= 0

a 2 + b 2 +c 2 = 38

A+b+c= 0 and a 2 + b 2 +c 2 = 38, find the value of a 3 +b 3 +c 3 -3abc  

Consider the formula - a³+b³+c³- 3abc = (a+b+c) (a²+b²+c²-(ab+bc+ca))

We have to find ab+bc+ca

given a+b+c = 0

Squaring on both sides we get,

(a+b+c)² = 0

a²+b²+c² + 2(ab+bc+ca) = 0

2 (ab+bc+ca) = -38

ab + bc + ca = -19

Now, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-(ab+bc+ca))

Putting the values we get

0 (38 - (-19))

0