Question

If x^2+1/x^2=7, find the value of x√3+1/x^3

Answer 1

Answer:

±18

Step-by-step explanation:

Given that,

${x}^{2} + \frac{1}{ {x}^{2} } = 7$

To find the value of :-

${x}^{3} + \frac{1}{ {x}^{3} }$

We know that,

$= > {x}^{2} + \frac{1}{ {x}^{2} } = 7 \\ \\ = > {(x)}^{2} + {( \frac{1}{x} )}^{2} + 2(x)( \frac{1}{x} ) = 7 + 2 \\ \\ = > {(x + \frac{1}{x} )}^{2} = 9 \\ \\ = > x + \frac{1}{x} = \pm \sqrt{9} \\ \\ = > x + \frac{1}{x} = \pm3$

Now, we know that,

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

Substituting the values, we get,

$= > {x}^{3} + \frac{1}{ {x}^{3} } = \pm3(7 - 1) \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = \pm3(6) \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = \pm18$

Hence, the required value is ±18.

Answer 2

CoRRᴇcᴛ Qᴜᴇsᴛɪᴏɴ :-

if (x² + 1/x²) = 7, find the value of (x³ + 1/x³) = ?

Sᴏʟᴜᴛɪᴏɴ :-

→ (x² + 1/x²) = 7

Adding 2 Both sides we get,

→ (x² + 1/x²) + 2 = 7 + 2

→ x² + 1/x² + 2 * x * 1/x = 9

Comparing the LHS part with a² + b² + 2ab = (a + b)² , we get,

→ (x + 1/x)² = 9

Square - Root Both sides now,

→ (x + 1/x) = ±3

________________

Now, if (x + 1/x) = +3

→ (x + 1/x) = 3

cubing Both sides,

→ (x + 1/x)³ = 3³

using (a + b)³ = a³ + b³ + 3ab(a + b) in LHS now,

→ x³ + 1/x³ + 3 * x * 1/x * (x + 1/x) = 27

→ x³ + 1/x³ + 3 * (x + 1/x) = 27

Putting value of (x + 1/x) Now,

→ (x³ + 1/x³) + 3 * 3 = 27

→ (x³ + 1/x³) + 9 = 27

→ (x³ + 1/x³) = 27 - 9

→ (x³ + 1/x³) = 18 (Ans.)

___________________

Now, if (x + 1/x) = (-3)

→ (x + 1/x) = (-3)

cubing Both sides,

→ (x + 1/x)³ = (-3)³

using (a + b)³ = a³ + b³ + 3ab(a + b) in LHS now,

→ x³ + 1/x³ + 3 * x * 1/x * (x + 1/x) = (-27)

→ x³ + 1/x³ + 3 * (x + 1/x) = (-27)

Putting value of (x + 1/x) Now,

→ (x³ + 1/x³) + 3 * (-3) = (-27)

→ (x³ + 1/x³) - 9 = (-27)

→ (x³ + 1/x³) = (-27) + 9

→ (x³ + 1/x³) = (-18) (Ans.)

__________________________