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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