Question

Find the value of a³+b³÷a³-b³​

Answer 1

Answer:

Required value of $\frac{a^{3}+b^{3}}{a^{3}-b^{3}}=\frac{9}{4\sqrt{5}}$

Step-by-step explanation:

Given,

$a=\frac{\sqrt{5}+1}{\sqrt{5}-1 }$   and   $b=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

Now,

$a+b=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}\\ =\frac{(\sqrt{5}+1)^{2}+(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}-1)}\\ =\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{5-1}\\ =\frac{12}{4}\\ =3$

and,

$a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}\\ =\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{(\sqrt{5}+1)(\sqrt{5}-1)}\\ =\frac{(5+1+2\sqrt{5})-(5+1-2\sqrt{5})}{5-1}\\ =\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}\\ =\frac{4\sqrt{5}}{4}\\ =\sqrt{5}$

and,

$ab=[\frac{\sqrt{5}+1}{\sqrt{5}-1}]*[\frac{\sqrt{5}-1}{\sqrt{5}+1}]\\ =\frac{(\sqrt{5}+1)(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}\\ =\frac{5-1}{5-1}\\ =\frac{4}{4}\\ =1$

∴$\frac{a^{3}+b^{3}}{a^{3}-b^{3}}\\ =\frac{(a+b)^{3}-3ab(a+b)}{(a-b)^{3}+3ab(a-b)}$

Putting the values of $(a+b),ab,(a-b)$ we get,

$\frac{3^{3}-3*3 }{(\sqrt{5})^{3}+3\sqrt{5}}\\ =\frac{27-9}{5\sqrt{5}+3\sqrt{5}}\\ =\frac{18}{8\sqrt{5}}\\ =\frac{9}{4\sqrt{5}}$

∴$\frac{a^{3}+b^{3}}{a^{3}-b^{3}}=\frac{9}{4\sqrt{5}}$

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Answer 2

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