Question

Find value of:- a) 8x^3 + y^3 - 30xy+ 125, if 2x+y=-5

Answer 1

Answer:

A) 2x+y=-5 ------------------------(1)

cubing both sides,

8x³+y³+3.2x.y(2x+y)=-125; [(x+y)³=x³+y³+3xy(x+y)]

or, 8x³+y³+6xy(-5)=-125 ; [using (1)]

or, 8x³+y³-30xy+125=0

b) x=3y-3 --------------------------(2)

cubing both sides,

x³=27y³-27-3.3y.3(3y-3) [(x-y)³=x³-y³-3xy(x-y)]

or, x³=27y³-27-27y(x) ; [using (2)]

or, x³-27y³+27=-27xy

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Answer 2

$Given \: (2x+y) = -5 \: ---(1)$

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We know the Algebraic Identity ;

(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)

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$Here, a = 2x \: and \: b = y$

$(2x+y)^{3} = (2x)^{3} + y^{3} + 3\times (2x)\times y ( 2x+y )$

$\implies (-5)^{3} = 8x^{3} + y^{3} + 6xy \times(-5) \:[From \:(1) ]$

$\implies -125 = 8x^{3} + y^{3} -30xy$

$\implies 0 = 8x^{3} + y^{3} -30xy+125$

Therefore.,

$\red{ Value \:of \: 8x^{3} + y^{3} -30xy+125}\green {= 0 }$

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