Question

if x^2+1/x^2 =146 find x^3-1/x^3

Answer 1

x²+1/x² = 146

(x)² + (1/x)² = 146

x²+(1/x)²+ 2(x)(1/x) - 2(x)(1/x) = 146

(x-1/x)²+2 = 146

(x-1/x)² = 146-2

x-1/x = √144

x-1/x = 12

(a³-b³)= (a-b)(a²+ab+b²)

(x³-(1/x)³) = (x-1/x)[x²+(x)(1/x)+(1/x)²]

→ (x³-1/x³) = (12)(x²+1+1/x²)

→ (x³-1/x³) = (12)(146+1) {x²+1/x² = 146}

→ (x³-1/x³) = 12(147) = 1764

Hope it helps

Answer 2

Answer:

Value of $x^3-\frac{1}{x^3}\:is\:1692$

Step-by-step explanation:

Given: $x^2+\frac{1}{x^2}=146$

To find: $x^3-\frac{1}{x^3}$

We use the identity,

( a - b )² = a² + b² - 2ab

put a = x and b = 1/x

we have,

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

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

$(x-\frac{1}{x})^2=146-2$

$x-\frac{1}{x}=\sqrt{144}$

$x-\frac{1}{x}=12$

Now, we use (a - b)³ = a³ - b³ -3ab( a - b )

put a = x  and b = 1/x

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

$(12)^3=x^3-\frac{1}{x^3}-3(12)$

$1728=x^3-\frac{1}{x^3}-36$

$x^3-\frac{1}{x^3}=1728-36$

$x^3-\frac{1}{x^3}=1692$

Therefore, Value of $x^3-\frac{1}{x^3}\:is\:1692$