\frac{x^3+y^3}{x^3-y^3}

ICSE Class X Math 2014 Solved Paper - Question 24

If $\;{x^2+y^2}/{x^2-y^2}=\;{17}/{8}$ , then find the value of:

Question : 24 of 52

Marks: +1, -0

\;{x^3+y^3}/{x^3-y^3}

Solution:

\;{x}/{y}=\;{5}/{3}

Cubing both sides, we get

\;{x^3}/{y^3}=\;{(5)^3}/{(3)^3}=\;{125}/{27}

Applying componendo and Dividendo

\;\;{x^3+y^3}/{x^3-y^3}=\;{125+27}/{125-27}

⇒ \;\; \;{x^3+y^3}/{x^3-y^3}=\;{152}/{98}

⇒ \;\; \;{x^3+y^3}/{x^3-y^3}=\;{76}/{49}

Go to Question:

Prev Question

Next Question