Question

If a+b=10 and a²+b²=58 find the value of a³+b³

Answer 1

(a+b)=10

a²+b²= 58

We know that

(a+b)²=a²+b²+2ab

(10)²=58+2ab

100=58+2ab

100-58= 2ab

42= 2ab

21= ab

Now ,

(a+b)³=a³+b³+3ab(a+b)

(10)³=a³+b³+3*21(10)

1000=a³+b³+63(10)

1000=a³+b³+630

1000-630=a³+b³

370=a³+b³

@:-)

Answer 2

(a+b)^2=10^2
a^2+2ab+b^2=100 (substitute values)
58+2ab=100
2ab=100-58
ab=42/2=21
Then
(a+b)^3=100^3
a^3+3ab(a+b)+b^3=1000000(then substitute values)
a^3+b^3+21 (10)=1000000
a^3+b^3=1000000-210
a^3+b^3=999,790