Simplifying expression, is it possible?

Question

To show that, $$ (7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3} = 2 $$ I am aware of the way where we can "guess" and come up with the following:

7 + 50^(1/2) = (1 + 2^(1/2))^3
7 - 50^(1/2) = (1 - 2^(1/2))^3

Hence simplifying the expression. But can we do it without the guesswork?

Answer 1

Hint:

Let $a,b=(7\pm\sqrt{50})^{1/3}$

$ab=-1$

$a^3+b^3=14$

Use $(a+b)^3=a^3+b^3+3ab(a+b)=14+3(-1)(a+b)$

So, $a+b$ is a real root of $$t^3+3t-14=0$$

whose only real root is $2$

Answer 2

Let $M = (7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3}$.

Then $M^3 = [(7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3}]^3=$

$(7+ 50^{\frac 12}) + 3(7 + 50^{1/2})^{2/3}(7 - 50^{1/2})^{1/3} + 3(7 + 50^{1/2})^{1/3}(7 - 50^{1/2})^{2/3} + (7-50^{\frac 12}) =$

$(7+ 50^{\frac 12})+ (7-50^{\frac 12})$$ + 3[(7+50^{\frac 12})(7-50^{\frac 12})]^{\frac 13}(7 + 50^{1/2})^{\frac 13} + 3[(7+50^{\frac 12})(7-50^{\frac 12})]^{\frac 13}(7 - 50^{1/2})^{\frac 13}=$

$14 +3(7^2 - (50^{\frac 12})^2)^{\frac 13})(7 + 50^{1/2})^{\frac 13} + 3(7^2 - (50^{\frac 12})^2)^{\frac 13})(7 - 50^{1/2})^{\frac 13}=$

$14 +3(49-50)^{\frac 13}[(7 + 50^{1/2})^{\frac 13}+(7 - 50^{1/2})^{\frac 13}]=$

$14 +3(-1)^{\frac 13}[(7 + 50^{1/2})^{\frac 13}+(7 - 50^{1/2})^{\frac 13}]=$

$14 - 3[(7 + 50^{1/2})^{\frac 13}+(7 - 50^{1/2})^{\frac 13}]=$

$=14 - 3M$.

So $(7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3}$ is a real solution to $M^3 +3M - 14=0$.

And if $M=2$ then $M^3 +3M -14 = 8+6-14=0$ so $M=2$ is also a real solution to $M^3 + 3M -14=0$.

How many real solutions does $M^3 + 3M -14=0$ have? And is it possible for $M=(7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3}$ and $M= 2$ to be two different solutions?

As $M =2$ is a solution, $M-2$ is a factor of $M^3 +3M -14$ and

$M^3 + 3M -14 = M^2(M-2) + 2M^2 + 3M -14 =$

$M^2(M-2) + 2M(M-2) + 4M + 3M -14 =$

$M^2(M-2) + 2M(M-2) + 7(M -2) + 14 -14 =$

$M^2(M-2) + 2M(M-2) + 7(M-2) = (M-2)(M^2 + 2M + 7)$.

If there are any other real solutions they will be solutions to $M^2 + 2M +7=0$ or

$M^2 + 2M + 1 = -6$ or

$(M+1)^2 = -6$. But if $M$ is real then $(M+1)^2 \ge 0$ so $M = 2$ is the only real solution.

And $(7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3}$ is a real number that is a solution.

So the only option is if $(7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3}$ and $2$ are the same number.

Answer 3

We can use also the following identity. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$

Let $a=\sqrt[3]{7-5\sqrt2},$ $b=\sqrt[3]{7+5\sqrt2}$ and $c=\sqrt[3]{7-5\sqrt2}+\sqrt[3]{7+5\sqrt2}$.

Thus, since $a+b-c=0,$ we obtain: $$7-5\sqrt2+7+5\sqrt2-c^3+3\sqrt[3]{(7-5\sqrt2)(7+5\sqrt2)}c=0$$ or $$c^3+3c-14=0$$ or $$c^3-2c^2+2c^2-4c+7c-14=0$$ or $$(c-2)(c^2+2c+7)=0$$ or $$c=2$$ and we are done!