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  • Find the value of (i) x^{3}+y^{3}-12xy-64 when x+y=-4 (ii) x^{3}-8y^{3}-36xy-216 when x=2y+6

    Find the value of

(i) x^{3}+y^{3}-12xy-64 when x+y=-4

(ii) x^{3}-8y^{3}-36xy-216 when x=2y+6

 

Answers (1)

(i) 0

Solution :-

Here x+y=-4

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc           

Here,

x+y+4=0

So using the above identity, we get:

x^{3}+y^{3}+4^{3}=3 \times x \times y\times 4= 12xy \cdots \cdots (i)

Now

\\x^{3}+y^{3}-12xy+64=x^{3}+y^{3}+(4)^{3}-12xy\\ 12xy-12xy=0{from equation i}

Hence the answer is 0

(ii) 0

Solution :-

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc           

Here,

x-2y-6=0

So using the above identity, we get:

x^{3}+(-2y)^{3}-(-6)^{3}=3x(-2y)(-6)= 36xy

x^{3}+8y^{3}-216=36xy                                           ....................(i)

Now

\\x^{3}-8y^{3}-216-36xy=36xy-36xy =0

Hence the answer is 0.

Posted by

infoexpert24

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