Question

x^2+1/x^2 =23 எ‌னி‌ல் x+1/x ம‌ற்று‌ம் x^3+1/x^3 ஆ‌கியவ‌ற்‌றி‌ன் ம‌தி‌ப்பை‌க் கா‌ண்க

Answer 1

||✪✪ QUESTION ✪✪||

if x² + 1/x² = 23 , find the value of x+1/x and x³ + 1/x³ ?

|| ✰✰ ANSWER ✰✰ ||

→ x² + 1/x² = 23

Adding 2 both sides we get,

→ x² + 1/x² + 2 = 23 + 2

Now, LHS Part can be written as ,

→ x² + 1/x² + 2 * x * 1/x = 25

Comparing LHS, with a² + b² + 2 * a * b now, we get,

→ (x + 1/x)² = 25

Square Root both sides now, we get,

→ (x + 1/x) = ±5 (Ans).

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Now, Lets take (x + 1/x) = 5 First ,

→ ( x + 1/x) = 5

Cubing both sides now, we get,

→ (x + 1/x)³ = 5³

Now, using (a+b)³ = a³ + b³ + 3ab(a+b) in LHS, we get,

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

→ x³ + 1/x³ + 3 * (x + 1/x) = 125

putting value of (x + 1/x) = 5 now, we get,

→ x³ + 1/x³ + 3 * 5 = 125

→ x³ + 1/x³ = 125 - 15

→ x³ + 1/x³ = 110 (Ans).

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Now, Lets Take (x + 1/x) = (-5)

→ ( x + 1/x) = (-5)³

Cubing both sides now, we get,

→ (x + 1/x)³ = (-5)³

Now, using (a+b)³ = a³ + b³ + 3ab(a+b) in LHS, we get,

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

→ x³ + 1/x³ + 3 * (x + 1/x) = (-125)

putting value of (x + 1/x) = (- 5) now, we get,

→ x³ + 1/x³ + 3 *(- 5) = (-125)

→ x³ + 1/x³ = (-125) + 15

→ x³ + 1/x³ = (-110) (Ans).

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Answer 2

விளக்கம்:

கொடுக்கப்பட்டுள்ள மதிப்பு

$x^{2}+\frac{1}{x^{2}}=23$ எனில் $x+\frac{1}{x}$ ம‌ற்று‌ம் ${x^3}+{\frac{1}{x^3} }$ ன் மதிப்பு

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

$\left(x+\frac{1}{x}\right)^{2}=23+2=25$

$\left(x+\frac{1}{x}\right)=\pm 5$

$x+\frac{1}{x}=5$

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

            $=5^{3}-3(5)$

            $=125-15$

            $=110$

  $x+\frac{1}{x}=-5$

$x^{3}+\frac{1}{x^{3}}=(-5)^{3}-3(-5)$

            $=-125+15$

            $=-110$

 ∴ $x+\frac{1}{x}=\pm 5, x^{3}+\frac{1}{x^{3}}=\pm 110$.