Please solve the question given in the attachment.
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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
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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 .
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