Question

X2+1/x2=27 then find the value of 3x3+5x-3/x3-5/x

Answer 1

Given :-

• x² + 1/x² = 27 .

To Find :-

• (3x³ + 5x - 3/x³ - 5/x) = ?

Solution :-

→ x² + 1/x² = 27

subtracting 2 from both sides,

→ x² + 1/x² - 2 = 27 - 2

→ (x)² + (1/x)² - 2 * x * 1/x = 25

→ (x - 1/x)² = 25

→ (x - 1/x) = ± 5

so, taking + 5

→ (x - 1/x)³ = (5)³

→ x³ - 1/x³ - 3 * x * 1/x(x - 1/x) = 125

→ x³ - 1/x³ - 3 * 5 = 125

→ (x³ - 1/x³) = 125 + 15

→ (x³ - 1/x³) = 140

and, taking (-5)

→ (x - 1/x)³ = (-5)³

→ x³ - 1/x³ - 3 * x * 1/x(x - 1/x) = (-125)

→ x³ - 1/x³ - 3 * (-5) = (-125)

→ (x³ - 1/x³) = (-125) - 15

→ (x³ - 1/x³) = (-140)

then,

→ (3x³ + 5x - 3/x³ - 5/x)

→ 3(x³ - 1/x³) + 5(x - 1/x)

→ 3 * 140 + 5 * 5

→ 420 + 25

→ 445 (Ans.)

or,

→ (3x³ + 5x - 3/x³ - 5/x)

→ 3(x³ - 1/x³) + 5(x - 1/x)

→ 3 * (-140) + 5 * (-5)

→ (-420) + (-25)

→ (-445) (Ans.)

Learn more :-

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– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

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Answer 2

Answer:

445

Step-by-step explanation:

x^2 + 1/x^2 = 27

(x - 1/x)^2 = x^2 + 1/x^2 - 2×x×1/x

ie. (x - 1/x)^2 = 27 - 2 = 25

So. (x - 1/x) = 5

(x - 1/x)^3 = x^3 - 1/x^3 - 3×x×1/x(x - 1/x)

5^3 = x^3 - 1/x^3 - 3×5

125 = x^3 - 1/x^3 - 15

x^3 - 1/x^3 = 125 + 15

x^3 - 1/x^3 = 140 . So

3x^3 + 5x - 3/x^3 - 5/x

= 3(x^3 - 1/x^3) +5(x - 1/x)

= 3 × 140 + 5×5

=420 + 25 = 445