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
(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 $27 x^{3}+8 y^{3}=7440$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.