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Prove It · #3 November 7th, 2009, 04:52 AM

Quote:

Originally Posted by deltaxray

Can someone please help me with this question.

A) if $x=2^{1/3} + 4^{1/3}$ show that $x^3 = 6(1+x)$

B) If $x= \frac {1}{2} + \frac {1}{2} \sqrt{5}$ , show that $\frac {x^2+x^{-2}}{x-x^{-1}} = 3$

Thanks in advance

$x^3 = \left(2^{\frac{1}{3}} + 4^{\frac{1}{3}}\right)^3$

$= \left(2^{\frac{1}{3}}\right)^3 + 3\left(2^{\frac{1}{3}}\right)^2\left(4^{\frac{1}{3}}\right) + 3\left(2^{\frac{1}{3}}\right)\left(4^{\frac{1}{3}}\right)^2 + \left(4^{\frac{1}{3}}\right)^3$

$= 2 + 3\cdot 2^{\frac{2}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}\cdot 4^{\frac{2}{3}} + 4$

$= 6 + 3\cdot 2^{\frac{4}{3}} + 3\cdot 2^{\frac{5}{3}}$

$6(1 + x) = 6\left(1 + 2^{\frac{1}{3}} + 4^{\frac{1}{3}}\right)$

$= 6 + 6\cdot 2^{\frac{1}{3}} + 6\cdot 2^{\frac{2}{3}}$

$= 6 + 3\cdot 2 \cdot 2^{\frac{1}{3}} + 3 \cdot 2 \cdot 2^{\frac{2}{3}}$

$= 6 + 3\cdot 2^{\frac{4}{3}} + 3\cdot 2^{\frac{5}{3}}$

$= x^3$ .