Question : If $x^3=270+y^3$ and $x=(6+y)$, then what is the value of $(x+y)? $(given that $x>0$ and $y>0$)
Option 1: $2 \sqrt{3}$
Option 2: $\sqrt{3}$
Option 3: $4 \sqrt{3}$
Option 4: $4 \sqrt{2}$
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Correct Answer: $4 \sqrt{3}$
Solution :
Given: $x^3=270+y^3$ and $x=(6+y)$
Putting the value of the $x=(6+y)$, we have:
⇒ $(6)^{3}+y^{3}+3×6×y(6+y)=270+y^{3}$
⇒ $216+y^{3}+18y(6+y)=270+y^{3}$
⇒ $108y+18y^{2}–54=0$
⇒ $y^{2}+6y–3=0$
The given quadratic equation is $y^{2}+6y-3=0$.
$y = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$
⇒ $y = \frac{-6 \pm \sqrt{6^{2}-4(1)(-3)}}{2(1)}$
⇒ $y = \frac{-6 \pm \sqrt{48}}{2}$
⇒ $y = \frac{-6 \pm 4\sqrt{3}}{2}$
⇒ $y = -3 \pm 2\sqrt{3}$
⇒ $y = -3 + 2\sqrt{3}$ ($\because y >$ 0)
We have, $x=(6+y)$
Putting the value, we get:
$x=6+(-3 \pm 2\sqrt{3})$
⇒ $x=3+ 2\sqrt{3}$
So, $(x+y)=-3 + 2\sqrt{3}+3+ 2\sqrt{3}$
$=4\sqrt{3}$
Hence, the correct answer is $4\sqrt{3}$.
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