Question

x^3+1\x^3 If the value of x is 7-4√3​

Answer 1

$\mathfrak{\huge{Answer:}}$

We are just given the value of x = $\sf{7 - 4\sqrt3}$

To simplify the calculation, what we can do is that find the value of $\tt{x + \frac{1}{x}}$ and then put the identity of $\sf{(a + b)^{3}}$ Where, a = x, and b = $\sf{\frac{1}{x}}$.

=》 $\tt{x + \frac{1}{x}}$ = $\sf{7 - 4\sqrt3 + \frac{1}{7 - 4\sqrt{3}}}$

After rationalizing, we find that :-

=》 $\tt{x + \frac{1}{x}}$ = $\sf{7 - 4 \sqrt{3} + 7 + 4 \sqrt{3}}$

Simplify it further

=》 $\tt{x + \frac{1}{x}}$ = $\sf{7 + 7}$

We get the value as

=》 $\tt{x + \frac{1}{x}}$ = 14

Now, put the identity of $\sf{(a + b)^{3}}$ Where, a = x, and b = $\sf{\frac{1}{x}}$

Solve this formed equation further

=》 $\tt{(x + \frac{1}{x})^{3}}$ = $\sf{x^{3} + \frac{1}{x} ^{3} + 3 ( x + \frac{1}{x})}$

Since, we have the value of $\tt{x + \frac{1}{x}}$ as 14, we can write the above equation as :-

=》 $\tt{14^{3}}$ = $\sf{x^{3} + \frac{1}{x} ^{3} + 3 (14)}$

Solve it further

=》 $\tt{2744}$ = $\sf{x^{3} + \frac{1}{x} ^{3} + 42}$

We get the final answer as

=》 $\tt{x^{3} + \frac{1}{x} ^{3} }$ = $\bf{2702}$

Answer 2

Answer :

Given :

x = 7 - 4√3

So we can say that,

$\sf{ \frac{1}{x} = \frac{1}{7 - 4 \sqrt{3}} }$

Rationalizing the denominator,

$\sf{\frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3}} }$

${\sf {= \frac{7 + 4 \sqrt{3} }{(7 + 4 \sqrt{3}) (7 - 4 \sqrt{3}) }} }$

Using the identity,

( a + b ) ( a - b ) = a^2 - b^2

$\sf{= \frac{7 + 4 \sqrt{3} }{(7) {}^{2} - (4 \sqrt{3 } ) {}^{2} }}$

$\tt{= 7 + 4 \sqrt{3}}$

Thus,

1/x =7+4√3

Now,

$\small \sf{x + \frac{1}{x} =( 7 - 4 \sqrt{3}) +( 7 + 4 \sqrt{ 3}})$

$\tt{ = 14}$

Now,

x^3 + 1/x^3 = (x + 1/x)^3 - 3 (1+ 1/x)

= (14)^3 - 3 × 14

= 2744 - 42

= 2702

Hence,

Answer to your question is 2702.