Question

Simplify: ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc$

Answer 1

Hint: Here the given first two terms are square of trinomial. To simplify the equation use the formula for square of trinomial $(a + b + c)$.The formula for square of trinomial is given by ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac$. So using this formula simplifies the first two terms of the above equation and after then arrange the terms and simplify the terms to obtain the final simplified form of the equation.

Complete step by step solution:Here the given equation is ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc$. We have to simplify the given equation. Here the given first two terms are square of trinomial. To simplify the equation we will use the formula for square of trinomial $(a + b + c)$.The formula for square of trinomial is given by ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac$.
So, let’s simplify the first term of the equation by using the formula for square of trinomial,
${(a + 2b + 3c)^2} = \left( {{a^2} + {{(2b)}^2} + {{(3c)}^2} + 2(a)(2b) + 2(2b)(3c) + 2(a)(3c)} \right)$
So, ${(a + 2b + 3c)^2} = {a^2} + 4{b^2} + 9{c^2} + 4ab + 12bc + 6ac$
Similarly let’s simplify the second term of the equation by using the formula for square of trinomial,
${(a - 2b + 3c)^2} = \left( {{a^2} + {{( - 2b)}^2} + {{(3c)}^2} + 2(a)( - 2b) + 2( - 2b)(3c) + 2(a)(3c)} \right)$
So, ${(a - 2b + 3c)^2} = \left( {{a^2} + 4{b^2} + 9{c^2} + ( - 4ab) + ( - 12bc) + 6ac} \right)$
Simplifying, ${(a - 2b + 3c)^2} = \left( {{a^2} + 4{b^2} + 9{c^2} - 4ab - 12bc + 6ac} \right)$
Now putting the expanded form of the above two terms in the main equation,
So, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc = ({a^2} + 4{b^2} + 9{c^2} + 4ab + 12bc + 6ac) - ({a^2} + 4{b^2} + {(3c)^2} - 4ab - 12bc + 6ac) - 6{b^2} - 9bc$Simplifying, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc = {a^2} + 4{b^2} + 9{c^2} + 4ab + 12bc + 6ac - {a^2} - 4{b^2} - 9{c^2} + 4ab + 12bc - 6ac - 6{b^2} - 9bc$
Cancelling the same but opposite sign terms,
So, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc = 4ab + 12bc + 4ab + 12bc - 6{b^2} - 9bc$
Simplifying, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc = 8ab + 15bc - 6{b^2}$
So, the simplified form of the equation ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc$ is $8ab + 15bc - 6{b^2}$.

Note: There is an alternate method to simplify the above equation. As the first two terms are in the form of ${a^2} - {b^2}$, we can use the formula as ${a^2} - {b^2} = (a - b)(a + b)$.
So let’s simplify the first two terms by using ${a^2} - {b^2} = (a - b)(a + b)$ formula,
So, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = \left\{ {(a + 2b + 3c) + (a - 2b + 3c)} \right\}\left\{ {(a + 2b + 3c) - (a - 2b + 3c)} \right\}$
Simplifying the terms, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = \left\{ {a + 2b + 3c + a - 2b + 3c} \right\}\left\{ {a + 2b + 3c - a + 2b - 3c} \right\}$
Cancelling the same but opposite sign terms, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = \left\{ {a + 3c + a + 3c} \right\}\left\{ {2b + 2b} \right\}$
So, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = \left\{ {2a + 6c} \right\}\left\{ {4b} \right\}$
Further simplifying terms, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = 2a(4b) + 6c(4b)$
So, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = 8ab + 24bc$
Now putting the value of ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} = 8ab + 24bc$in main equation,
So, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc = 8ab + 24bc - 6{b^2} - 9bc$
Simplifying, ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc = 8ab + 15bc - 6{b^2}$
So the simplified form of the equation ${(a + 2b + 3c)^2} - {(a - 2b + 3c)^2} - 6{b^2} - 9bc$ is $8ab + 15bc - 6{b^2}$.