Question

cube root a + cube root b +cube root c = 0 then what is (a+b+c)whole cube ?

Answer 1

Answer:

$(a+b+c)^3=27abc$

Step-by-step explanation:

Formula used:

if x+y+z=0 then $x^3+y^3+z^3=3xyz$

Given:

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

From the above formula

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

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

Now,

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

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

$(a+b+c)^3=27abc$

Answer 2

Answer:

$\text{The value of }(a+b+c)^3\text{ is }27abc$

Step-by-step explanation:

Given that

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

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

As we know

$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3abc$

If x+y+z=0, then

$x^3+y^3+z^3=3xyz$

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

Then, we get

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

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

Take cube on both sides

$(a+b+c)^3=27(\sqrt[3]{a})^3(\sqrt[3]{b})^3(\sqrt[3]{c})^3=9abc$

$\text{Hence, the value of }(a+b+c)^3\text{ is }27abc$