If x+(1/x)=7, then x3-(1/x3) is

Question

If x+(1/x)=7, then x3-(1/x3) is (A) 9 √5 (B) 144 √5 (C) 135 √5 (D) √5

Tardigrade Admin · Accepted Answer

(i) Use the formulae (a+b)2 and (a-b)3. (ii) Find x-(1/x) by using (x-(1/x))2-4=(x-(1/x))2. (iii) Find x3-(1/2 x3) by using (x-(1/x))3 +3 x ⋅((1/x))(x-(1/x))

Q. If $x + \frac{1}{x} = 7$ , then $x^{3} - \frac{1}{x ^{3}}$ is

$95$

$1445$

$1355$

$5$

(i) Use the formulae $(a + b)^{2}$ and $(a - b)^{3}$ .
(ii) Find $x - \frac{1}{x}$ by using $(x - \frac{1}{x})^{2} - 4 = (x - \frac{1}{x})^{2}$ .
(iii) Find $x^{3} - \frac{1}{2 x ^{3}}$ by using $(x - \frac{1}{x})^{3}$
$+ 3 x \cdot (\frac{1}{x}) (x - \frac{1}{x})$

Q. If x+x1​=7, then x3−x31​ is

95​

1445​

1355​

5​

(i) Use the formulae (a+b)2 and (a−b)3. (ii) Find x−x1​ by using (x−x1​)2−4=(x−x1​)2. (iii) Find x3−2x31​ by using (x−x1​)3 +3x⋅(x1​)(x−x1​)

Q. If $x + \frac{1}{x} = 7$ , then $x^{3} - \frac{1}{x ^{3}}$ is

$95$

$1445$

$1355$

$5$

(i) Use the formulae $(a + b)^{2}$ and $(a - b)^{3}$ .
(ii) Find $x - \frac{1}{x}$ by using $(x - \frac{1}{x})^{2} - 4 = (x - \frac{1}{x})^{2}$ .
(iii) Find $x^{3} - \frac{1}{2 x ^{3}}$ by using $(x - \frac{1}{x})^{3}$
$+ 3 x \cdot (\frac{1}{x}) (x - \frac{1}{x})$