What is expand (1+1+1=0,)(1+2+3=0) ?

The solution to expand (1+1+1=0,)(1+2+3=0) is 6=0

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expand (1+1+1=0,)(1+2+3=0)

expand

(1 + 1 + 1 = 0,) (1 + 2 + 3 = 0)

6 = 0

Solution steps

(1 + 1 + 1 = 0,) (1 + 2 + 3 = 0)

= (1 + 1 + 1 \cdot 0 =,) (1 + 2 + 3 \cdot 0 =)

Distribute parentheses

1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 0 + 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 0 + 1 = 0, 1 + 1 = 0, 2 + 1 = 0, 3 = 0

Expand

1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 : 6

6 = 0

52 (x^{'} + h (x))^{2}

\frac{1}{x ^{2} ( x + a )}

A (3, - 2) B (3, - 6)

\frac{( 2 x + 3 ) ^{2}}{2}

(1 + 2 e^{2 t})^{2}

What is expand (1+1+1=0,)(1+2+3=0) ?
The solution to expand (1+1+1=0,)(1+2+3=0) is 6=0