यदि $4a~+\frac{4}{a}~+~5~=~0$, तो $({{a}^{3}}~+\frac{1}{{{a}^{3}}}~-\frac{5}{2})$ का मान ज्ञात करें |
$\frac{25}{32}$
$-\frac{25}{64}$
$\frac{45}{32}$
$-\frac{45}{64}$
Solution
$4a~+\frac{4}{a}~+~5~=~0$
$4a~+\frac{4}{a}~=~-5$
$\left( a~+\frac{1}{a} \right)~=~-5/4$
अब $\left( a~+\frac{1}{a} \right)3~=~-\frac{125}{64}$
${{a}^{3}}~+\frac{1}{{{a}^{3}}~}+~3\times a\times \left( \frac{1}{a}
\right)\left( a+\frac{1}{a} \right)~=~-\frac{125}{64}$
${{a}^{3}}~+\frac{1}{{{a}^{3}}}~-\frac{15}{4}~=~-\frac{125}{64}$
${{a}^{3}}~+\frac{1}{{{a}^{3}}}~=~-\frac{125}{64}~+\frac{15}{4}~=\frac{115}{64}$
अतः ,
${{a}^{3~}}+\frac{1}{{{a}^{3}}}~-\frac{5}{2}~=\frac{115}{64}~-\frac{5}{2}~=~-\frac{45}{64}$