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Algebraic Identities

If m-1m=5, ...

Question

If $m - \frac{1}{m} = 5$ , then find $m^{4} + \frac{1}{m^{4}}$

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Solution

Applying the formula, ${(a - b)}^{2} = a^{2} + b^{2} - 2 a b$
${(m - \frac{1}{m})}^{2} = m^{2} + {(\frac{1}{m})}^{2} - 2 m \times \frac{1}{m} = m^{2} + \frac{1}{m^{2}} - 2$
$=> m^{2} + \frac{1}{m^{2}} = {(m - \frac{1}{m})}^{2} + 2$
Substituting $m - \frac{1}{m} = 5$ ,
$=> m^{2} + \frac{1}{m^{2}} = 5^{2} + 2 = 25 + 2 = 27$
Squaring both sides,
${(m^{2} + \frac{1}{m^{2}})}^{2} = 27 \times 27 = 729$
$m^{4} + \frac{1}{m^{4}} + 2 \times m^{2} \times (\frac{1}{m^{2}}) = 729$
$m^{4} + \frac{1}{m^{4}} + 2 = 729$
$m^{4} + \frac{1}{m^{4}} = 727$

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Similar questions

If $m - \frac{1}{m} = 5,$ find :

(i) $m^{2} + \frac{1}{m^{2}}$

(ii) $m^{4} + \frac{1}{m^{4}}$

(iii) $m^{2} - \frac{1}{m^{2}}$

m = 2

, then find the value of

m^{5} \div m^{6} \times {(\frac{1}{m})}^{4}

m^{5} + m^{4} + m^{3} + m^{2} + m + 1 = (m^{3} + 1) \times

Q. For the system shown in

6.339, m_{1} > m_{2} > m_{3} > m_{4}

. Initially, the system is at rest in equilibrium condition. If the string joining

m_{4}

and ground is cut, then just after the string is cut
Statement I:

m_{1}, m_{2}, m_{3}

remain stationary.
Statement II: The value of the acceleration of all the four blocks can be determined.
Statement III: Only

m_{4}

remains stationary.
Statement IV: Only

m_{4}

accelerates.
Now, choose the correct options.

31 m 4

is divisible by

3

, then what is the value of

m

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