Question

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

Answer 1

It is given that x + y + 4 = 0

we have to find the value of (x³ + y³ - 12xy + 64)

here x + y + 4 = 0

⇒x + y = -4 ........(1)

we know, a³ + b³ = (a + b)³ - 3ab(a + b)

so, x³ + y³ = (x + y)³ - 3xy(x + y)

= (-4)³ - 3xy(-4) [ from equation (1) ]

= -64 + 12xy .........(2)

now x³ + y³ - 12xy + 64 = -64 + 12xy - 12xy - 64 = 0

therefore value of x³ + y³ - 12xy + 64 = 0

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Answer 2

Answer:

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Step-by-step explanation:

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