Question #191706

Find the value of a³+2a²b+b³ divided by ab(a+3b) when a/b=√2-1


1
Expert's answer
2021-05-17T04:02:02-0400

We have given that,


ab=21\dfrac{a}{b} = \sqrt{2}-1


a=b(21)a = b(\sqrt{2}-1)

We have to find the value of a3+2ab+b3ab(a+3b)\dfrac{a^3+2a^b+b^3}{ab(a+3b)}


Putting the value of a in the above equation,


We get,


b3(21)3+2b3(21)2+b3b3(21)2+3b3(21)\dfrac{b^3(\sqrt{2}-1)^3+2b^3(\sqrt{2}-1)^2+b^3}{b^3(\sqrt{2}-1)^2+3b^3(\sqrt{2}-1)}


=(21)3+(21)2+1(21)2+(21)= \dfrac{(\sqrt{2}-1)^3+(\sqrt{2}-1)^2+1}{(\sqrt{2}-1)^2+(\sqrt{2}-1)}


After solving we get,


32122\dfrac{3 \sqrt{2}-1}{2-\sqrt{2}}


=32= \dfrac{3}{\sqrt{2}}


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