Question 4

What is the value of $$\frac{a^2+b^2}{a^3-b^3}$$ when $$a+b=8$$ and $$a-b=2$$

Solution

Given : $$a+b=8$$ and $$a-b=2$$

Adding both equations, => $$2a=8+2=10$$

=> $$a=\frac{10}{2}=5$$

Similarly, $$b=8-5=3$$

To find : $$\frac{a^2+b^2}{a^3-b^3}$$

= $$\frac{(5)^2+(3)^2}{(5)^3-(3)^3}$$

= $$\frac{25+9}{125-27}=\frac{34}{98}\approx0.347$$

=> Ans - (B)

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