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Find the value of:
[tex] 27x^{3}+8 y^{3} [/tex] if 3x+2y = 20 and xy = [tex] \frac{14}{9} [/tex]

Answer :

[tex]27x^3+8y^3=(3x)^3+(2y)^3=(3x+2y)[(3x)^2-3x\cdot2y+(2y)^2]\\\\=(3x+2y)[(3x)^2+2\cdot3x\cdot2y+(2y)^2-3\cdot3x\cdot2y]\\\\=(3x+2y)[(3x+2y)^2-18xy]=(*)\\---------------------------\\3x+2y=20\\\\xy=\frac{14}{9}\\------------------------------\\(*)=20\cdot(20^2-19\cdot\frac{14}{9})=20\cdot(400-18\cdot\frac{14}{9})=20\cdot(400-2\cdot14)\\\\=20\cdot(400-28)=20\cdot372=7440[/tex]



[tex]a^3+b^3=(a+b)(a^2-ab+b^2)\\\\(a+b)^2=a^2+2ab+b^2[/tex]

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