x+1/x=4; x^3+1/x^3=?
after making a quadratic equation of this, which btw doesn't factorize, I am stuck in the square root of 12. Also, is there any other way to find the value of x or the answer?
+26400 x+1/x=4;
x^3+1/x^3=?
\(\begin{array}{rcll} x+ \frac{1}{x} &=& 4 \qquad & | \qquad ()^3\\ \left( x+ \frac{1}{x} \right)^3 &=& 4^3 \\ x^3 + 3\cdot x^2\cdot \frac{1}{x} + 3\cdot x \cdot \frac{1}{x^2} + \frac{1}{x^3} &=& 64 \\ x^3 + 3\cdot x + \frac{3}{x} + \frac{1}{x^3} &=& 64 \\ x^3 + \frac{1}{x^3} + 3\cdot x + \frac{3}{x} &=& 64 \\ x^3 + \frac{1}{x^3} + 3\cdot \left( x + \frac{1}{x} \right) &=& 64 \qquad & | \qquad x+ \frac{1}{x} = 4\\ x^3 + \frac{1}{x^3} + 3\cdot 4 &=& 64 \\ x^3 + \frac{1}{x^3} + 12 &=& 64 \qquad & | \qquad -12\\ x^3 + \frac{1}{x^3} &=& 64 -12\\\\ \mathbf{ x^3 + \frac{1}{x^3} } &\mathbf{=}& \mathbf{52}\\ \end{array}\)
+26400 x+1/x=4;
x^3+1/x^3=?
\(\begin{array}{rcll} x+ \frac{1}{x} &=& 4 \qquad & | \qquad ()^3\\ \left( x+ \frac{1}{x} \right)^3 &=& 4^3 \\ x^3 + 3\cdot x^2\cdot \frac{1}{x} + 3\cdot x \cdot \frac{1}{x^2} + \frac{1}{x^3} &=& 64 \\ x^3 + 3\cdot x + \frac{3}{x} + \frac{1}{x^3} &=& 64 \\ x^3 + \frac{1}{x^3} + 3\cdot x + \frac{3}{x} &=& 64 \\ x^3 + \frac{1}{x^3} + 3\cdot \left( x + \frac{1}{x} \right) &=& 64 \qquad & | \qquad x+ \frac{1}{x} = 4\\ x^3 + \frac{1}{x^3} + 3\cdot 4 &=& 64 \\ x^3 + \frac{1}{x^3} + 12 &=& 64 \qquad & | \qquad -12\\ x^3 + \frac{1}{x^3} &=& 64 -12\\\\ \mathbf{ x^3 + \frac{1}{x^3} } &\mathbf{=}& \mathbf{52}\\ \end{array}\)
