please solve this question tomorrow is my exam. its urgent
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3x³/81-2y³/128-x²y/12+xy²/16
=x³/27-y³/64-x²y/12+xy²/16
=(x/3)³-(y/4)³-xy/4(x/3+y/4)
=(x/3-y/4)³ [(a-b)³=a³-b³-3ab(a+b)]
=x³/27-y³/64-x²y/12+xy²/16
=(x/3)³-(y/4)³-xy/4(x/3+y/4)
=(x/3-y/4)³ [(a-b)³=a³-b³-3ab(a+b)]
Ankit1234:
thanks
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3x^3/81 - 2y^3/128 - x^2y/12 + xy^2/16
identity will be used (a-b)^3 = a^3 - b^3 -3ab^2 + 3a^2b
3x^3 2y^3 x^2 y xy^2
=> _______ - ______ - ______ + _____
3x3x3x3 2x64 12 16
x^3 y^3 x^2 y xy^2
= > _____ - ____ - ______ + _____
27 64 12 16
=> x y
{ __ }^3 - {___}^3 - (3)(x/3)^2(y/4) + (3)(x/3)(y/4)^2
3 4
=>(x/3 -y/4)^3
identity will be used (a-b)^3 = a^3 - b^3 -3ab^2 + 3a^2b
3x^3 2y^3 x^2 y xy^2
=> _______ - ______ - ______ + _____
3x3x3x3 2x64 12 16
x^3 y^3 x^2 y xy^2
= > _____ - ____ - ______ + _____
27 64 12 16
=> x y
{ __ }^3 - {___}^3 - (3)(x/3)^2(y/4) + (3)(x/3)(y/4)^2
3 4
=>(x/3 -y/4)^3
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