# How to solve: 2a * b ^ 2 * c – 2a + 3b ^ 3 * c – 3b – 4b ^ 2 * c ^ 2 + 4c

Welcome to my article How to solve: 2a * b ^ 2 * c – 2a + 3b ^ 3 * c – 3b – 4b ^ 2 * c ^ 2 + 4c This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve: 2a * b ^ 2 * c – 2a + 3b ^ 3 * c – 3b – 4b ^ 2 * c ^ 2 + 4c read and understand it carefully till the end.

Let us know how to solve this question How to solve: 2a * b ^ 2 * c – 2a + 3b ^ 3 * c – 3b – 4b ^ 2 * c ^ 2 + 4c.

First write the question on the page of the notebook.

## How to solve: 2a * b ^ 2 * c – 2a + 3b ^ 3 * c – 3b – 4b ^ 2 * c ^ 2 + 4c

On writing this query like this,

\displaystyle 2a{{b}^{2}}c-2a+3{{b}^{3}}c-3b-4{{b}^{2}}{{c}^{2}}+4c

\displaystyle 2a\times {{b}^{2}}\times c-2a+3{{b}^{3}}\times c-3b-4{{b}^{2}}\times {{c}^{2}}+4c

\displaystyle 2a{{b}^{2}}c-2a+3{{b}^{3}}c-3b-4{{b}^{2}}{{c}^{2}}+4c

\displaystyle 2a\times {{b}^{2}}c-2a\times 1+3b\times {{b}^{2}}c-3b\times 1-4c\times {{b}^{2}}c+4c\times 1

\displaystyle 2a({{b}^{2}}c-\times 1)+3b\times {{b}^{2}}c-3b\times 1-4c\times {{b}^{2}}c+4c\times 1

\displaystyle 2a({{b}^{2}}c-1)+3b({{b}^{2}}c-1)-4c\times {{b}^{2}}c+4c\times 1

\displaystyle 2a({{b}^{2}}c-1)+3b({{b}^{2}}c-1)-4c({{b}^{2}}c-1)