Question

4c(-a+b+c) - [3a(a+b+c)-2b(a-b+c)

Answer 1

(-4ac+4bc+4c^2)-[3a^2+3ab+3ac-2ab+2b^2-2bc]

-4ac+4bc+4c^2-3a^2-3ab-3ac+2ab-2b^2+2bc

-7ac +6bc-ab+4c^2-3a^2-2b^2

Answer 2

Question

$\sf{4c(-a + b + c)-[3a(a + b +c) -2b(a + b + c)]}$

Solution

$\sf{ - 4ac + 4bc + {4c}^{2} - [3 {a}^{2} + 3ab + 3ac - 2ab + {2b}^{2} - 2bc ]}$

$\sf{4ac + 4bc + {4c}^{2} - [3 {a}^{2} + 2 {b}^{2} + 3ab - 2bc + 3ac - 2ab]}$

$\sf{4ac + 4bc + {4c}^{2} - [3 {a}^{2} + 2 {b}^{2} + ab + 3ac - 2bc]}$

= 4ac + 4bc + 4c² + 3a² - 2b² - ab - 3ac + 2bc

= - 3a² - 2b² + 4c² - ab + 4bc + 2bc - 4ac - 3ac

= 3a² - 2b² + 4c³ + ab + 6bc - 7ac

Hence,Solved

$\begin{gathered}\boxed{\begin{array}{cc}\boxed{\star\:\:\textsf{\textbf{Algebric\:Identity}}\:\star}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\sf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\sf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\sf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}\end{gathered}$