Question

Find the value of x^3 + y^3 - 12abc + 64,when x+y = 4.​

Answer 1

Answer:

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

Step-by-step explanation:

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Answer 2

solution is...............