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Question

Question 39
Find the value of:
(i)
x3+y312xy+64, when x + y = - 4

(ii) x38y336xy216, when x = 2y + 6

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Solution

(i) Given
x + y + 4 = 0
x3+y3+(4)3=3xy(4) …….(1)
[Using the identity, if a+b+c=0, then a3+b3+c3=3abc]
=12xy
Now,
x3+y312xy+64=x3+y3+6412xy
=12xy12xy=0 [from Eq. (i) ]

(ii) Given,
x2y6=0
x3+(2y)3+(6)3=3x(2y)(6)
[Using the identity, if a+b+c=0, then a3+b3+c3=3abc]
x38y3216=36xy …..(i)
Now, x38y336xy216
=x38y321636xy
=36xy36xy=0 [from Eq. (i)]

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