If the value of m4+1m4=119, then the value of |m3−1m3|=

m4+1m4=119⇒(m2)2+1m22=119Add ‘2’ on both sides⇒(m2)2+1m22+2=119+2⇒m2+1m22=121m2+1m2=11Subtract ‘2’ on both sidesm2+1m2−2=11−2m−1m2=9⇒m−1m=3|m3−1m3|=|m-1mm2+1m2+1|=|3×12|=36

If the value of m4+1m4=119, then the value of |m3−1m3|=

m4+1m4=119⇒(m2)2+1m22=119Add ‘2’ on both sides⇒(m2)2+1m22+2=119+2⇒m2+1m22=121m2+1m2=11Subtract ‘2’ on both sidesm2+1m2−2=11−2m−1m2=9⇒m−1m=3|m3−1m3|=|m-1mm2+1m2+1|=|3×12|=36

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If the value of $m^{4} + \frac{1}{m^{4}} = 119$ , then the value of $| m^{3} - \frac{1}{m^{3}} | =$

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Detailed Solution

$\begin{array}{l} m^{4} + \frac{1}{m^{4}} = 119 \\ \Rightarrow {(m^{2})}^{2} + {(\frac{1}{m^{2}})}^{2} = 119 \end{array}$

Add ‘2’ on both sides

$\begin{array}{l} \Rightarrow {(m^{2})}^{2} + {(\frac{1}{m^{2}})}^{2} + 2 = 119 + 2 \\ \Rightarrow {(m^{2} + \frac{1}{m^{2}})}^{2} = 121 \\ m^{2} + \frac{1}{m^{2}} = 11 \end{array}$

Subtract ‘2’ on both sides

$\begin{array}{l} m^{2} + \frac{1}{m^{2}} - 2 = 11 - 2 \\ {(m - \frac{1}{m})}^{2} = 9 \Rightarrow m - \frac{1}{m} = 3 \\ | m^{3} - \frac{1}{m^{3}} | = | (m - (\frac{1}{m})) (m^{2} + \frac{1}{m^{2}} + 1) | \\ = | 3 \times 12 | = 36 \end{array}$