Question

(a+3c)²+(b+3c)(b-3c)​

Answer 1

Ответ:

Ответ смотри на фото...

Answer 2

Ответ:

$a^{2}+6ac+b^{2}$

Объяснение:

$(a+b)^{2}=a^{2}+2ab+b^{2}, \quad (a+b)(a-b)=a^{2}-b^{2};$

$(a+3c)^{2}+(b+3c)(b-3c)=a^{2}+2 \cdot a \cdot 3c+(3c)^{2}+b^{2}-(3c)^{2}=a^{2}+6ac+b^{2};$

Answer 3

To expand the expression $(a+3c)^2 + (b+3c)(b-3c)$, you can use the distributive property and then apply the formula for squaring a binomial. Here's how you can do it step by step:

1. Expand $(a+3c)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

   $(a+3c)^2 = a^2 + 2(a)(3c) + (3c)^2$

   Simplify:

   $a^2 + 6ac + 9c^2$

2. Expand $(b+3c)(b-3c)$ using the difference of squares formula $a^2 - b^2 = (a+b)(a-b)$:

   $(b+3c)(b-3c) = b^2 - (3c)^2$

   Simplify:

   $b^2 - 9c^2$

Now, you can combine the two simplified expressions:

$(a+3c)^2 + (b+3c)(b-3c) = (a^2 + 6ac + 9c^2) + (b^2 - 9c^2)$

Now, you can add the like terms together:

$a^2 + 6ac + 9c^2 + b^2 - 9c^2$

Finally, simplify further if needed:

$a^2 + b^2 + 6ac - 9c^2 + 9c^2$

The $9c^2$ terms cancel out:

$a^2 + b^2 + 6ac$

So, the simplified expression is $a^2 + b^2 + 6ac$.