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I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
Put $a = \sqrt[3]{5x+7}, b = \sqrt[3]{5x-12}\implies a - b = 1, a^3 - b^3 =19\implies a^2+ab+b^2 = \dfrac{a^3-b^3}{a-b} = \dfrac{19}{1} = 19\implies (a-b)^2+3ab = 19\implies 1^2+3ab = 19 \implies ab = 6\implies (b+1)b = 6\implies b^2+b-6 = 0\implies (b+3)(b-2)=0\implies b = 2, -3\implies 5x-12 = 2^3, (-3)^3 = 8,-27\implies 5x = 20, -15\implies x = 4, -3.$
Before cubing, separate your roots:
$$\sqrt[3]{5x+7}=\sqrt[3]{5x-12}+1$$
Now cube:
$$5x + 7 = 5x - 12 + 3\sqrt[3]{5x-12}^2 + 3\sqrt[3]{5x-12} + 1$$
Simplify:
$$ \sqrt[3]{5x-12}^2 + \sqrt[3]{5x-12} - 6 = 0$$
Solve for $\sqrt[3]{5x-12}$ (it is a quadratic equation). Finally solve to $x$.
Let's see. We have
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$$
now set $x=y+\frac12$. Then
$$\sqrt[3]{5y+\tfrac{19}2}-\sqrt[3]{5y-\tfrac{19}2}=1$$
Now cubing yields
$$19-3\sqrt[3]{(5y)^2-(\tfrac{19}2)^2}\left(\sqrt[3]{5y+\tfrac{19}2}-\sqrt[3]{5y-\tfrac{19}2}\right)=1$$
which rewrites to
$$\sqrt[3]{(5y)^2-(\tfrac{19}2)^2}=6$$
Which is easy to solve, $25y^2=6^3+(\frac{19}2)^2$ or $y=\pm \frac 72$. Now we conclude $x=4$ or $x=-3$.
Following up on OP's "almost-there" attempt (and correcting the $-1\color{red}{2}\,$ in the second equation):
$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1 \tag{1}$
i tried to cube both sides, but i ended up with an equation looking like this:
$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1\color{red}{2}}-\sqrt[3]{5x+7})=-6 $
The rightmost factor is $-1$ per the original equation $(1)\,$, which leaves:
$$-\sqrt[3]{(5x-12)(5x+7)}=-6 \tag{2}$$
Then the numbers $\,\sqrt[3]{5x+7}\,$ and $\,-\sqrt[3]{5x-12}\,$ have sum $\,1\,$ per $(1)\,$, and product $\,-6\,$ per $(2)\,$, so they are the roots of the quadratic $\,t^2-t-6=0 \iff t \in \{-2, 3\}\,$:
$\sqrt[3]{5x+7} = -2 \iff 5x +7 = -8 \iff x = -3$
$\sqrt[3]{5x+7} = 3 \iff 5x + 7 = 27 \iff x = 4$
$$a=\sqrt[3]{5x+7},b=\sqrt[3]{5x-12}\\a-b=1\\(a-b)^3=1\\a^3-3a^2b+3ab^2-b^3=1\\a^3-b^3-3ab(a-b)=1\\a^3-b^3-3ab=1\\5x+7-5x+12-3\sqrt[3]{5x+7}\sqrt[3]{5x-12}=1\\-3\sqrt[3]{5x+7}\sqrt[3]{5x-12}=-18\\\sqrt[3]{5x+7}\sqrt[3]{5x-12}=6\\(5x+7)(5x-12)=216\\25x^2-60x+35x-84=216\\25x^2-25x-300=0\\x^2-x-12=0\\x_1=4,x_2=3$$
Actually, there is a nice, generic solution to equations of this type, namely equations in the form
$$\sqrt[3]{f(x)}+\sqrt[3]{g(x)} = C,$$
especially if $f(x)$ and $g(x)$ sum to a constant.
Cubing and rearranging, we get
$$f(x) + g(x) + 3C\sqrt[3]{f(x)g(x)}=C^3,$$
or
$$\sqrt[3]{f(x)g(x)} = \frac{C^3-f(x)-g(x)}{3C}.$$
Cubing again, we get
$$f(x)g(x) = \frac{(C^3-f(x)-g(x))^3}{27C^3}.$$
If $f(x)$ and $g(x)$ are linear in $x$, this is a cubic equation. But if $f(x)+g(x)=A$ const., we get
$$f(x)[A-f(x)] = \frac{(C^3-A)^3}{27C^3},$$
or
$$f(x)^2 - Af(x) + \left[\frac{(C^3-A)}{3C}\right]^3 = 0,$$
which is quadratic in $f(x)$.
In the specific example, $f(x)=5x+7$, $g(x)=-5x+12$, $A=f(x)+g(x)=19$, $C=1$, and we get
$$f(x)^2 - 19f(x) - 216=0,$$
which yields $f(x)=5x+7=27$ or $f(x)=5x+7=-8$, i.e., $x=4$ or $x=-3$.
