**If x**^{2} + 1/x^{2} = 27, find the value of 3x^{3} + 5x – 3/x^{3} – 5/x.

^{2}+ 1/x

^{2}= 27, find the value of 3x

^{3}+ 5x – 3/x

^{3}– 5/x.

**Answer :**

We know that

**(x – 1/x) ^{2} = x^{2} + 1/x^{2} – 2**

Substituting the values

(x – 1/x)^{2} = 27 – 2 = 25

So we get

x – 1/x = ± √25 = ± 5

Here

3x^{3} + 5x – 3/x^{3} – 5/x = 3 (x^{3} – 1/x^{3}) + 5 (x – 1/x)

It can be written as

= 3 [(x – 1/x)^{3} + 3 (x – 1/x)] + 5 (x – 1/x)

Substituting the values

= 3 [(± 5)^{3} + 3 (± 5)] + 5 (± 5)

By further calculation

= 3 [(± 125) + (± 15)] + (± 25)

So we get

= (± 375) + (± 45) + (± 25)

= ± 445

**More Solutions:**

- If (x + 1/x)2 = 3, find x3 + 1/x3.
- If x = 5 – 2√6, find the value of √x + 1/√x.
- If a + b + c = 12 and ab + bc + ca = 22, find a2 + b2 + c2.
- If a + b + c = 12 and a2 + b2 + c2 = 100, find ab + bc + ca.
- If a2 + b2 + c2 = 125 and ab + bc + ca = 50, find a + b + c.
- If a2 + b2 + c2 = 125 and ab + bc + ca = 50, find a + b + c.
- Find the value of ab – bc – ca.
- If a – b = 7 and a2 + b2 = 85, then find the value of a3 – b3.
- Find the product of x and y.
- Find the sum of their cubes.