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Question 116
If $$a^2+\frac{1}{a^2}$$ = 98(a > 0) then the value of $$a^3 + \frac{1}{a^3}$$ will be
Solution
$$(a+\frac{1}{a})^{2}$$=$$a^{2}+\frac{1}{a^{2}}+2$$=98+2=100
So $$(a+\frac{1}{a})$$=10
or $$(a+\frac{1}{a})^{3}$$=1000
or $$a^{3}+\frac{1}{a^{3}}$$+3($$(a+\frac{1}{a})$$)=1000
or  $$a^{3}+\frac{1}{a^{3}}$$+$$3\times10=1000$$
or $$a^{3}+\frac{1}{a^{3}}$$=970
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