Question

if 2a-2/a+1=0 find the value of a3-1/a3+2=?

Answer 1

Answer:

${a}^{3} - \frac{1}{ {a}^{3} } + 2 = \frac{3}{8}$

Step-by-step explanation:

If

$2a - \frac{2}{a} + 1 = 0...eq1$

then to find the value of

${a}^{3} - \frac{1}{ {a}^{3} } + 2$

take cube of eq1

$2a - \frac{2}{a} = - 1 \\ \\ a - \frac{1}{a} = \frac{ - 1}{2} \\ \\ {(a - \frac{1}{a}) }^{3} = {( \frac{ - 1}{2} )}^{3} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3 {a}^{2} ( \frac{1}{a} ) + 3a( \frac{1}{ {a}^{2} } ) = - \frac{1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3a + \frac{3}{a} = \frac{ - 1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3(a - \frac{1}{a}) = \frac{ - 1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3( - \frac{1}{2}) = \frac{ - 1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } + \frac{3}{2} = \frac{ - 1}{8}$

${a}^{3} - \frac{1}{ {a}^{3} } + 2 = \frac{ - 1}{8} - \frac{3}{2} + 2 \\ \\ = \frac{ - 1 - 12 + 16}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } + 2 = \frac{3}{8}$

Hope it helps you.

Answer 2

Answer:

2a-2/a+1=0 than a3-1/a3+2 what answer step