Find the values of

Question:

Find the values of $27 x^{3}+8 y^{3}$, if

(a) 3x + 2y = 14 and xy = 8

(b) 3x + 2y = 20 and xy = 14/9

 

Solution:

(a) Given, 3x + 2y = 14 and xy = 8

cubing on both sides

$(3 x+2 y)^{3}=14^{3}$

We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

$27 x^{3}+8 y^{3}+3(3 x)(2 y)(3 x+2 y)=2744$

$27 x^{3}+8 y^{3}+18 x y(3 x+2 y)=2744$

$27 x^{3}+8 y^{3}+18(8)(14)=2744$

$27 x^{3}+8 y^{3}+2016=2744$

$27 x^{3}+8 y^{3}=2744-2016$

$27 x^{3}+8 y^{3}=728$

Hence, the value of $27 x^{3}+8 y^{3}=728$

(b) Given, 3x + 2y = 20 and xy = 14/9

cubing on both sides

$(3 x+2 y)^{3}=20^{3}$

We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

$27 x^{3}+8 y^{3}+3(3 x)(2 y)(3 x+2 y)=8000$

$27 x^{3}+8 y^{3}+18 x y(3 x+2 y)=8000$

$27 x^{3}+8 y^{3}+18(14 / 9)(20)=8000$

$27 x^{3}+8 y^{3}+560=8000$

$27 x^{3}+8 y^{3}=8000-560$

$27 x^{3}+8 y^{3}=7440$

Hence, the value of $27 x^{3}+8 y^{3}=7440$

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