Question 78

If $$a^3 + b^3 + c^3 = 8072, a : b = b : c = 3 : 2$$ then the value of a is

Solution

Let b=6m Then a=9m and c=4m

$$a^3+b^3+c^3=8072$$

$$\left(9m\right)^3+\left(6m\right)^3+\left(4m\right)^3=8072$$

$$729m^3+216m^3+64m^3=8072$$

$$1009m^3=8072\ or\ m^3=8\ or\ m=2$$

Now a=9m or a=18

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