Question

$a + \frac{4}{a } = - 2 \: then \: (a {3 - 8) = }^{?}$
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Answer 1

Answer:

0

Step-by-step explanation:

Given :

$a + \frac{4}{a } = - 2$

To Find :

${a}^{3} - 8$

Solution :

$a + \frac{4}{a } = - 2 \\ \\ \frac{ {a}^{2} + 4 }{a} = - 2 \\ \\ {a}^{2} + 4 = - 2a \\ \\ {a}^{2} + 2a + 4 = 0 \: \: \: \: \: \: \: .........i$

Again,

${a}^{3} - 8 \\ \\ ( {a}^{} - 2)( {a}^{2} + 2a + 4) \\ \\ (a - 2) \times 0 \: \: \: \: \: \: \:</u></strong><strong><u>[</u></strong><strong><u> from \: i </u></strong><strong><u>]</u></strong><strong><u>\\ \\ 0$

Using formula

a³ - b³ = (a - b)(a² + ab + b²)

Answer 2

Given:

$a + \frac{4}{a} = - 2$

To find:

${a}^{3} - 8$

Solution:

$a + \frac{4}{a} = - 2 \\ \\ (taking \: lcm \: in \: lhs) \\ \\ \frac{ {a}^{2} + 4 }{a} = - 2 \\ \\ (cross \: multiplying) \\ \\ {a}^{2} + 4 = - 2a \\ \\ {a}^{2} + 2a + 4 = 0 ----equation(1)-$

we have to find

${a}^{3} - 8 \\ \\ = {a}^{3} - {(2)}^{3} \\ \\ (using \: identity \\ \: {x}^{3} </em></strong><strong><em>-</em></strong><strong><em> {y}^{3} = (x </em></strong><strong><em>-</em></strong><strong><em> </em></strong><strong><em>y)( {x}^{2} + {y}^{2} </em></strong><strong><em>+</em></strong><strong><em> xy) \: we \: will \: get)$

$(a - 2)( {a}^{2} + 4 + 2a) \\ \\ (using \: equation \: (1) \: we \: will \: get) \\ \\</em></strong><strong><em>=</em></strong><strong><em> (a - 2)(0) \\ \\ = 0$

Answer

your answer will be 0

I. e,

${a}^{3} - 8 = 0$