Question

Please solve the question given in the attachment.

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Answer 1

Answer:

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

Given, a+b+c=0

∴a3+b3+c3−3abc=0

∴a3+b3+c3=3abc

Answer 2

a + b + c = 4

a² + b² + c² = 10 ( 4 + 6 = 10 )

a³ + b³ + c³ = 22 ( 10 + 12 = 22)

a⁴ + b⁴ + c⁴ = 40 ( 22 + 18 = 40 ) .

here the reason is 6 is added to 4 which is 6 × 1 while 12 is added to 10 which is 6 × 2 while 18 is added to 22 which is 6 × 3 .