If \(\rm k^4+\frac{1}{k^4}=194\), then what is the value of \(\rm k^3+\frac{1}{k^3}\)?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 02 Dec 2022 Shift 3)

Answer 1

Given:

\(\rm k^4+\frac{1}{k^4}=194\)

Concept used:

\(\rm (a+\frac{1}{a})^2=\rm a^2+\frac{1}{a^2} + 2\)

If

\(\rm (a+\frac{1}{a})=b\)

Then,

\(\rm a^3+\frac{1}{a^3} =b^3-3b\)

Calculation:

\(\rm k^4+\frac{1}{k^4}=194\)

⇒ \(\rm k^4+\frac{1}{k^4}+2=194+2\)

⇒ \(\rm k^4+\frac{1}{k^4}+2=196\)

⇒ \(\rm( k^2+\frac{1}{k^2})^2=14^2\)

⇒ \(\rm k^2+\frac{1}{k^2}=14\)

⇒ \(\rm k^2+\frac{1}{k^2}+2=14+2\)

⇒ \(\rm (k+\frac{1}{k})^2=16\)

⇒ \(\rm (k+\frac{1}{k})^2=4^2\)

⇒ \(\rm k+\frac{1}{k}=4\)

Now,

\(\rm k^3+\frac{1}{k^3}\) = 4³ - 3 × 4

⇒ \(\rm k^3+\frac{1}{k^3}\) = 64 - 12

⇒ \(\rm k^3+\frac{1}{k^3}\) = 52

∴ Ther required answer is 52.

Shortcut Trick

We know,

• If k⁴ + 1/k⁴ = a then k² + 1/k² = √(a + 2)

• If k2 + 1/k2 = b then k + 1/k = √(b + 2)

• If k + 1/k = c then k³ + 1/k³ = c³ - 3c

So, k4 + 1/k4 = 194 then k2 + 1/k2 = √196 = 14 so, k + 1/k = √16 = 4

Then, k3 + 1/k3 = 4³ - 12 = 52