x³+y³-12xy+64 Factorise this please....
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Step-by-step explanation:
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Find the value of x
3
+y
3
−12xy+64, when x+y=−4.
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Answer
Correct option is
A
0
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:
x ^ { 3 } + y ^ { 3 } - 12 x y + 64
Consider x^{3}+y^{3}-12xy+64 as a polynomial over variable x.
x^{3}-12yx+y^{3}+64
Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{3} and m divides the constant factor y^{3}+64. One such factor is x+y+4. Factor the polynomial by dividing it by this factor.
(x+ y+4) (x^{2}-xy-4x+y^{2}-4y+16)