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