Find "A" in this equation.

Question

$$ \sqrt[3] {A-15√3} + \sqrt[3] {A+15√3} = 4 $$

Find "A" ? The way of exponentiation took too much time, is there any easier method?

Answer 1

Let $x,y$ be the numbers $(A\pm 15\sqrt 3)^{1/3}$ so that $x+y=4$. hen we have: $$ \begin{aligned} 2A &= x^3+y^3\\ &= (x+y)^3-3xy(x+y) \\ &= 4^3-3xy\cdot 4 \\ &=64 - 3(A^2-675)^{1/3}\cdot 4\ .\text{ So:} \\[3mm] 6^3(A^2-675) &=(32-A)^3\ . \end{aligned} $$ This gives us an equation in $A$, that we may solve (with bare hands or not).

sage: var('A');
sage: solve( (32 - A)^3 - 6^3*(A^2-675) == 0, A )
[A == -9*I*sqrt(19) - 73, A == 9*I*sqrt(19) - 73, A == 26]

OK, $26$ is a solution (obtained with bare hands or not), the only real solution.

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LATER EDIT: It seems that there is a real interest in finding solutions with bare hands. Let us do this explicitly. The equation of degree three in $A$, $$ (A-32)^3+6^3(A^2-675) =0 $$ becomes after expanding: $$ A^{3} + 120 \, A^{2} + 3072 \, A - 178568 = 0\ . $$ To have smaller coefficients it is humanly useful to observe that $2|120$, $4|3072$, $8|178568$, so we may substitute $A=2B$, and get the equation in $B$: $$ h(B):= B^3 + 60B^2 + 768B - 22321 = 0\ . $$ If this equation has rational solutions (as psychologically expected in human problems), then any rational solution is of the form $\pm p/q$, where $p$ divides the free coefficient $-22321=13 \cdot 17 \cdot 101$, and $q$ divides the principal coefficient (of $B^3$), which is one. So we compute $h(0)<0$, expect a positive solution, and try the values $1, 13, 17, 13\cdot 17, 101, 13\cdot 101,\dots$ and note immediately that $h(100) >0$, so only $1,13,17$ may have a chance. The $1$ is not. The we try humanly $1$ with the Horner scheme, $$ \begin{array}{r|rrrr} &1 & 60 & 678 & -22321\\\hline 13 & 1 & 73 & 1717 & \boxed{0} \end{array} $$ where the numbers in the second row are $1$, copied from above, $73=13\cdot 1+60$, $1717=13\cdot 73+678$, and finally $\boxed{0}=13\cdot 1717-22321$.

So one solution is $B=13$ and we have the factorization $$ h(B):= B^3 + 60B^2 + 768B - 22321 =(B-13)(B^2+73B+1717)\ , $$ and the other two solutions are roots of $h(B)=0$ are the ones of $(B^2+73B+1717)=0$, explicitly $(-73\pm 9i\sqrt{19})/2$. To get the $A$ values, we multiply with two.

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So far we have only one direction. For the other direction, we check that the obtained values lead to third roots that we can explicitly express.

$$ \begin{aligned} (26\pm 15\sqrt 3)^{1/3} &= 2\pm \sqrt 3\ , \\ (-73+9\sqrt{-19}\pm 15\sqrt 3)^{1/3} &= 2 \mp \frac{\sqrt 3}2(1+\sqrt{-19})\ , \\ (-73-9\sqrt{-19}\pm 15\sqrt 3)^{1/3} &= 2 \mp \frac{\sqrt 3}2(1-\sqrt{-19})\ . \end{aligned} $$ But in the case of using complex numbers, the third root is not "defined in the common sense", and the "two other choices" would break the equality.

Answer 2

Let $x=A-15\sqrt 3$ and $y=A+15\sqrt 3,$ then the equation becomes $$x^{1/3}+y^{1/3}=4.$$ Taking cubes of both sides gives $$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$ Now since $x+y=2A,$ and $x^{1/3}+y^{1/3}=4,$ the equation becomes $$2A+3(xy)^{1/3}(4)=4^3,$$ which gives $$6(xy)^{1/3}=32-A.$$ Now, cubing and substituting for $xy=A^2-15^2\cdot 3$ gives $$6^3(A^2-15^2\cdot 3)=(32-A)^3.$$ This is a cubic in $A,$ which in general is not so easy to deal with.

Answer 3

If there exists a cubic polynomial of integer coefficients with roots: $$0,\sqrt[3] {A-15√3} , \sqrt[3] {A+15√3} $$ then it will be of the form $x^3-4x^2 + \sqrt[3]{A^2-675}x$

You just need to find $A$ such that $A^2-675$ is a perfect cube.
Trivially $A^2 = 676$ will do.