Proof that a*b*c=0 therefore a cube b cube c cube = 3abc


In this proof, we will show that if the product of three numbers a, b, and c is zero, then the sum of their cubes is equal to three times their product.


Let's start with the given information that abc = 0. We can rewrite this as:

a * b * c = 0

Now, we can consider the expression (a + b + c) cubed, which expands to:

a^3 + b^3 + c^3 + 3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) + 6abc

Notice that the first three terms on the right-hand side are the cubes of a, b, and c, respectively. We can rewrite the rest of the expression in terms of the product abc:

3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) = 3(ab + ac + bc)(a + b + c)

Now, if we substitute our given information abc = 0 into this expression, we get:

3(ab + ac + bc)(a + b + c) = 0

This means that either ab + ac + bc = 0 or a + b + c = 0.

If a + b + c = 0, then we can rewrite our original expression (a + b + c) cubed as:

(a + b + c) cubed = 3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2)

Substituting abc = 0, we get:

a^3 + b^3 + c^3 = 3abc

which is the result we wanted to prove.

If ab + ac + bc = 0, we can rewrite our original expression (a + b + c) cubed as:

(a + b + c) cubed = 3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) - 6abc

Substituting abc = 0 and ab + ac + bc = 0, we get:

a^3 + b^3 + c^3 = 3abc

which is again the result we wanted to prove.


We have shown that if the product of three numbers a, b, and c is zero, then the sum of their cubes is equal to three times their product. This is a useful result in many mathematical applications.

A^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
if a+b+c=0 then a^3+b^3+c^3-3abc=(0)(a^2+b^2+c^2-ab-bc-ca)
a^3+b^3+c^3-3abc=0
a^3+b^3+c^3=3abc