Question

if cube root of a + cube root of b + cube root of c = 0, then (a+b+c)^3= ?​

Answer 1

Answer:

Answer:

\text{The value of }(\frac{a+b+c}{3})^3\text{ is }abcThe value of (3a+b+c)3 is abc

Step-by-step explanation:

Given that

\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0\

\text{we have to find the value of }(a+b+c)^3we have to find the value of (a+b+c)3

As we know

x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3abcx3+y3+z3=(x+y+z)(x2+y2+z2−xy−yz−xz)+3abc

If x+y+z=0, then

x^3+y^3+z^3=3xyzx3+y3+z3=3xyz

Put\thinspace x=\sqrt[3]{a}, y=\sqrt[3]{b}, z=\sqrt[3]{c}\

Then, we get

(\sqrt[3]{a})^3+(\sqrt[3]{b})^3+(\sqrt[3]{c})^3=3(\sqrt[3]{a})(\sqrt[3]{b})(\sqrt[3]{c})(3a)3+(3b)3+(3c)3=3(3a)(3b)(3c)

⇒ a+b+c=3(\sqrt[3]{a})(\sqrt[3]{b})(\sqrt[3]{c})a+b+c=3(3a)(3b)(3c)

\frac{a+b+c}{3}=(\sqrt[3]{a})(\sqrt[3]{b})(\sqrt[3]{c})3a+b+c=(3a)(3b)(3c)

Take cube on both sides

(\frac{a+b+c}{3})^3=(\sqrt[3]{a})^3(\sqrt[3]{b})^3(\sqrt[3]{c})^3=9abc(3a+b+c)3=(3a)3(3b)3(3c)3=9abc

\text{Hence, the value of }(a+b+c)^3\text{ is }abcHence, the value of (a+b+c)3 is abc

Answer 2

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Hope it's helps you ✔