Find the value of a³+2a²b+b³ divided by ab(a+3b) when a/b=√2-1
We have given that,
ab=2−1\dfrac{a}{b} = \sqrt{2}-1ba=2−1
a=b(2−1)a = b(\sqrt{2}-1)a=b(2−1)
We have to find the value of a3+2ab+b3ab(a+3b)\dfrac{a^3+2a^b+b^3}{ab(a+3b)}ab(a+3b)a3+2ab+b3
Putting the value of a in the above equation,
We get,
b3(2−1)3+2b3(2−1)2+b3b3(2−1)2+3b3(2−1)\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)}b3(2−1)2+3b3(2−1)b3(2−1)3+2b3(2−1)2+b3
=(2−1)3+(2−1)2+1(2−1)2+(2−1)= \dfrac{(\sqrt{2}-1)^3+(\sqrt{2}-1)^2+1}{(\sqrt{2}-1)^2+(\sqrt{2}-1)}=(2−1)2+(2−1)(2−1)3+(2−1)2+1
After solving we get,
32−12−2\dfrac{3 \sqrt{2}-1}{2-\sqrt{2}}2−232−1
=32= \dfrac{3}{\sqrt{2}}=23
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