Question

If x+y+4=0 find the value of x^3+y^3-12xy+64

Answer 1

Answer:

a³+b ³+c³

=(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

Step-by-step explanation:

x+y+4=0

So, x+y = 0-4

= -4

x^3+y^3-12xy+64

So, x³+y³+64 = 4×3xy

= 12xy

Now,

x^3+y^3+64-12xy = 12xy - 12xy

So, x^3+y^3-12xy+64 = 0 Ans.