Question

find the value of x³+y³+12xy-64 when x+y=4​

Answer 1

We know that,

a

3

+b

3

+c

3

=(a+b+c)(a

2

+b

2

+c

2

−ab−bc−ca)+3abc

If a+b+c=0, then a

3

+b

3

+c

3

=3abc

Now, given x

3

+y

3

−12xy+64 and

x+y=−4

=>x+y+4=0

Here, a=x, b=y, c=4 and a+b+c=x+y+4=0

Therefore

x

3

+y

3

+64=3xy(4)

=12xyz

Now,

x

3

+y

3

+64−12xyz=12xyz−12xyz

=0

Answer 2

hope it will help you....