[2a³c-2a³b+2ab³-2b³c+2bc³-2ac³]
please I need step factoring that guy up there
2[ a³(c- b)+ b³( a-c)+ c³ (b- a)] by pulling out common factors
yes one call reiny did it...but I don't know how she did it....been trying these for days now help bro
To factor the expression [2a³c-2a³b+2ab³-2b³c+2bc³-2ac³], we can observe that it contains six terms. We will factor it by grouping the terms into pairs and then factoring out the common factors from each pair.
Step 1: Group the terms into pairs:
[2a³c - 2a³b] + [2ab³ - 2b³c] + [2bc³ - 2ac³]
Step 2: Factor out the common factors from each pair:
2a³c - 2a³b can be factored as 2a³(c - b)
2ab³ - 2b³c can be factored as 2b³(a - c)
2bc³ - 2ac³ can be factored as 2c³(b - a)
Now our expression becomes:
2a³(c - b) + 2b³(a - c) + 2c³(b - a)
Step 3: Rearrange the terms:
Rearranging the terms to make it look more organized, we get:
2a³(c - b) - 2b³(c - a) + 2c³(b - a)
Now we can observe that the expression contains three terms and each term has a common factor of (c - b). We can factor out (c - b) to further simplify the expression:
(c - b)(2a³ - 2b³ + 2c³)
The final factored form of the expression [2a³c-2a³b+2ab³-2b³c+2bc³-2ac³] is (c - b)(2a³ - 2b³ + 2c³).