Question

If X=2+√3, then find X cubed +1/X cubed
This is the written version of the question in the above given image.
Please solve this.
It is urgent.
I will mark you as brainliest.
please​

Answer 1

Gɪᴠᴇɴ :-

• x = 2 + √3

ᴛᴏ ꜰɪɴᴅ :-

• (x³ + 1/x³)

Sᴏʟᴜᴛɪᴏɴ :-

→ x = 2 + √3

→ 1/x = 1/(2+√3)

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

→ 1/x = (2 - √3) / ((2 + √3)(2 - √3)

→ 1/x = (2 - √3) / (2² - (√3)²) [(a + b)(a - b) = (a² - b²)]

→ 1/x = (2 - √3) / ( 4 - 3)

→ 1/x = (2 - √3)

So,

→ (x + 1/x) = (2 + √3) + (2 - √3)

→ (x + 1/x) = 4

Cubing both sides now we get,

→ (x + 1/x)³ = 4³

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

→ x³ + 1/x³ + 3 * x * 1/x(x + 1/x) = 64

→ (x³ + 1/x³) + 3 * 4 = 64

→ (x³ + 1/x³) = 64 - 12

→ (x³ + 1/x³) = 52 (Ans.)

Answer 2

Answer:

⋆ Given : x = 2 + √3

⋆ we will find the value of $\sf\frac{1}{x}$

• whenever two continuous numbers are given in roots, then just put Opposite sign.
• That will be Reciprocal of it.

$:\implies\sf x=2+\sqrt{3}\\\\\\:\implies\sf x = \sqrt{4} + \sqrt{3}\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Reciprocal of x will be :}}\\\\:\implies\sf \dfrac{1}{x} = \sqrt{4} - \sqrt{3}\\\\\\:\implies\sf \dfrac{1}{x} =2 - \sqrt{3}$

$\rule{150}{1}$

$\underline{\bigstar\:\textbf{According to the Question :}}$

$\dashrightarrow\sf\:\:x + \dfrac{1}{x} =2 + \sqrt{3} + 2 - \sqrt{3}\\\\\\\dashrightarrow\sf\:\:x + \dfrac{1}{x} = 4\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{\bigg\lgroup x+\dfrac{1}{x}=a\bigg\rgroup\quad then\quad \bigg\lgroup x^3+\dfrac{1}{x^3}=(a)^3-3a\bigg\rgroup}}\\\\\dashrightarrow\sf\:\:x^3+\dfrac{1}{x^3} =(4)^3 - (3 \times 4)\\\\\\\dashrightarrow\sf\:\:x^3+\dfrac{1}{x^3} = 64 - 12\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf x^3+\dfrac{1}{x^3} = 52}}$

⠀

$\therefore\:\underline{\textsf{Required value of the given will be \textbf{52}}}.$