Math, asked by kavya1011, 1 year ago

Find the value of a³+b³+6ab-8 when a+b=2

Answers

Answered by Anonymous
34
a3+b3+6ab-8= (a+b)3-3ab(a+b)+6ab-8

=(2)3-3ab*2+6ab-8

=8-6ab+6ab-8

=0
Answered by 9552688731
68
(a+b) = 2
cube both side
(a+b)³ = (2)³
a³ + b³ + 3ab(a+b) = 8
a³ + b³ + 3ab(2). = 8
a³ + b³ + 6ab. = 8
a³ + b³ + 6ab - 8. = 0

so the value of a³+b³+6ab -8. is. 0
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