Question

Given ,
x=1 +root2 +root3 then find (x-1)³ +1/(x-1)³​

Answer 1

Gɪᴠᴇɴ :-

• x = (1 + √2 + √3)

Tᴏ Fɪɴᴅ :-

• (x-1)³ +1/(x-1)³

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• a³ + b³ = a³ + b³ + 3ab(a + b)

Sᴏʟᴜᴛɪᴏɴ :-

→ x = (1 + √2 + √3)

→ (x - 1) = (√2 + √3) -------- Eqn.(1)

→ 1/(x - 1) = 1/(√2 + √3)

→ 1/(x - 1) = 1/(√2 + √3) * {(√2 - √3) / (√2 - √3)}

→ 1/(x - 1) = (√2 - √3) / {(√2)² - (√3)²}

→ 1/(x - 1) = (√2 - √3)/(2 - 3)

→ 1/(x - 1) = -(√2 - √3) = (√3 - √2) -------- Eqn(2) .

Now,

→ (x - 1) + 1/(x - 1) = √2 + √3 + √3 - √2

→ (x - 1) + 1/(x - 1) = 2√3

cubing Both sides,

→ [(x - 1) + 1/(x - 1)]³ = (2√3)³

→ (x - 1)³ + 1/(x - 1)³ + 3*(x-1)*1/(x-1)[ (x - 1) + 1/(x - 1) ] = 8*3√3

→ (x - 1)³ + 1/(x - 1)³ + 3 * 1 * 2√3 = 24√3

→ (x - 1)³ + 1/(x - 1)³ + 6√3 = 24√3

→ (x - 1)³ + 1/(x - 1)³ = 24√3 - 6√3

→ (x - 1)³ + 1/(x - 1)³ = 18√3 (Ans.)

Answer 2

★ Given:-

x = (1+√2+√3)

✶ To Find :-

${x - 1}^{3} + \frac{1}{ {x - 1}^{3} }$

FoRMula To solVe the MAth :-

{a^3+b^3 =a^3+b^3 +3ab(a+b)

$\large\mathtt{SOLution:-}$

$\implies \: x = (1 + \sqrt{2} + \sqrt{3}$

$\implies \: (x - 1) = ( \sqrt{2} + \sqrt{3} ) - - - - - - - (i)$

$\implies \: \frac{1}{x - 1} = \frac{1}{ \sqrt{2 } + \sqrt{3} }$

$\implies \: \frac{1}{x - 1} = \sqrt{2} + \sqrt{3} \frac{{ \sqrt{2} }^{2} }{ \sqrt{3 {?}^{2} } }$

$\implies \: \frac{1}{x - 1} = - ( \sqrt{2} - \sqrt{3} = \sqrt{3 - \sqrt{2} } ) - - - - - - - (ii)$

NoW,

$\implies \: (x - 1) + \frac{1}{x - 1} \sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2}$

$\implies \: (x - 1) + \frac{1}{x - 1} = \sqrt[2]{3}$

Therefore,

Cubing the both sides,

⇒(x-1) ^3 +(1) (x-1)^3+3*(x-1) * 1/(x-1) [(x-1) +1/(x-1)] =8*3√3

⇒ (x-1)^3 + (1)(x-1)^3.1.2√3 =24√3

⇒(x-1)^3 + (1)(x-1)^3 +6√3 =24√3

⇒ {x-1}^3 + 1/(x-3) ^3=24√3-6√3

⇒ {x-1}^3 + = 18√3

$\large\mathtt{BE BRAINLY}$