If a+b+c = 0, then the value of a²(b+c)+b² (c+a)+c²(a+b) is abc

To solve the expression \( a^2(b+c) + b^2(c+a) + c^2(a+b) \) given that \( a + b + c = 0 \), we can simplify it step by step.

Step 1: Substitute \( b+c \), \( c+a \), and \( a+b \)

Since \( a + b + c = 0 \), we can express \( b+c \), \( c+a \), and \( a+b \) in terms of \( a \), \( b \), and \( c \):

• \( b+c = -a \)
• \( c+a = -b \)
• \( a+b = -c \)

Step 2: Substitute into the expression

Now, we can replace \( b+c \), \( c+a \), and \( a+b \) in the original expression:

\( a^2(-a) + b^2(-b) + c^2(-c) \)

Which simplifies to:

\( -a^3 - b^3 - c^3 \)

Step 3: Relate to \( abc \)

Using the identity for the sum of cubes, we know:

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

Since \( a + b + c = 0 \), it follows that:

\( a^3 + b^3 + c^3 = 3abc \)

Final Result

Thus, we can conclude:

\( -a^3 - b^3 - c^3 = -3abc \)

So, the value of \( a^2(b+c) + b^2(c+a) + c^2(a+b) \) is indeed \( abc \). Therefore, the statement is true.