Question

if x-y = 15 and xy = 20 then find the value of 1/x³ - 1/y³​

Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• x-y = 15
• x*y = 20

To find:

• 1/x³ - 1/y³

Solution:

$\displaystyle \implies \dfrac{1}{x^3} - \dfrac{1}{y^3} \\ \\ \implies \dfrac{y^3}{x^3*y^3} - \dfrac{x^3}{x^3*y^3} \\ \\ \implies \dfrac{y^3-x^3}{(x*y)^3} \\ \\ \text{We know that a^3-b^3 = (a-b)(a^2+ab+b^2) So, } \\ \\ \implies \dfrac{(y-x)(y^2 + xy + x^2)}{(x*y)^3} \\ \\ \text{We know that a^2 + b^2 = (a-b)^2 + 2ab . So,} \\ \\ \implies \dfrac{(y-x)*(xy + (y-x)^2 + 2xy)}{(xy)^3} \\ \\ Putting\:values\:of\:(y-x)=-15\:\&\:xy=20:- \\ \\ \implies \dfrac{(-15)(20+(-15)^2+2*20)}{20^3} \\ \\ \implies \dfrac{(-15)*(20+40+225)}{8000} \\ \\ \implies \dfrac{-15*285}{8000} \\ \\ \implies \dfrac{-3*57}{320} (dividing\:both\:sides\: by\: 25) \\ \\ \implies \dfrac{-171}{320}.$

Formula used:

• $a^3-b^3 = (a-b)(a^2+ab+b^2)$
• $a^2 + b^2 = (a-b)^2 + 2ab$

Learn more:

• $a^3-b^3 = (a-b)(a^2+ab+b^2)$
• $a^3+b^3 = (a+b)(a^2-ab+b^2)$
• $(a+b)^2 = a^2 + 2ab + b^2$
• $(a-b)^2 = a^2 - 2ab + b^2$
• $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
• $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
• $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
• $a^2 - b^2 = (a+b)*(a-b)$

Answer 2

ANSWER :

X - Y = 15 AND XY = 20

THEN 1 / X CUBE - 1 / Y CUBE

= - 69 / 90