Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.

Question

Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.

I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.

p.s. The answers are $\pm 6\sqrt 3$.

try cubing after taking one part with 2.like $a+b=2 \implies a=2-b$ then cube — iostream007, May 15 '13 at 09:59
$\pm 6\sqrt{3}$ is the answer for $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$ — Isomorphism, May 15 '13 at 10:16
@Maazul, we need to take sufficient care while editing Questions. You have changed sign . — lab bhattacharjee, May 15 '13 at 10:45

score 6 · Accepted Answer · answered May 15 '13 at 10:29

6

Set $x+10=t^3$ and $x-10 = u^3$. So we want to solve $t - u = 2$.

However $t^3 - u^3 = (t-u)^3 + 3tu(t-u) = 20$.

That is $2^3 + 6tu = 8 + 6tu =20$.

Now we have $tu = 2$. This means $x^2 -100 = 8$ and thus $x = \pm6\sqrt{3}.$

answered May 15 '13 at 10:29

Isomorphism

$t+u=2$ right? Where did you get $t-u=2?$ – lab bhattacharjee May 15 '13 at 10:41
@labbhattacharjee: I think the OP edited the question. Now it is $t-u = 2$. Isnt it? – Isomorphism May 15 '13 at 10:43
It wasn't OP who changed the question, but it looks like it arose in the comments. – Thomas Andrews May 16 '13 at 12:54

score 4 · Answer 2 · answered May 15 '13 at 10:27

$\displaystyle\sqrt[3]{x+10}-\sqrt[3]{x-10}=2---(1)$

Let $y=x+10$

$(1)$ becomes

$\displaystyle\sqrt[3]{y}-\sqrt[3]{y-20}=2$

$\displaystyle\sqrt[3]{y}=2+\sqrt[3]{y-20}$

Cube both sides to get

$\displaystyle{y}=8+12 (y-20)^{\frac{1}{3}}+6(y-20)^{\frac{2}{3}}+y-20$

$\displaystyle{0}=-12+ 6 (y-20)^{\frac{1}{3}}(2+(y-20)^{\frac{1}{3}})$

$\displaystyle{12}= 6 (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$

$\displaystyle{2}= (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$

Cube both sides

$\displaystyle{8}= (y-20)(y)$

$\displaystyle{y^2-20y-8}= 0$

Solving the above equation yields $y=10\pm 6\sqrt{3}$ which implies $x=\pm 6\sqrt{3}$.

2 Answers2