The coefficient of $x^3$ in the expansion of $(1+2x)^6(1-x)^7$ is 

$\textbf{Step 1: Expand the Expression }$

$\text{Binomial Theorem is Given by-}$

$(1+p)^n=1+^nC_{1}(p)^1+^nC_{2}(p)^2+^nC_{3}(p)^3+..............+^nC_{n}(p)^n$

$\text{For expansion of} (1+2x)^6:$

$\text{By Comparing with Formula p=2x and n=6}$

$(1+2x)^6 =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\}$

$\text{For expansion of} (1-x)^7 :$

$\text{By Comparing with Formula p=x and n=7}$

$(1-x)^7 = \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..........\}$

$\textbf{Step 2: Find the coefficient of } \mathbf{ x^3}$

$\text{Let P}= (1+2x)^6(1-x)^7$

$\Rightarrow P =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\} \times \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..\}$

$\therefore \text{Coefficient of }x^3 = -^{7}C_3 + ^{7}C_2 \times 2 ^{6}C_1 + 4\times ^{6}C_2 \times (- ^{7}C_1 )+8\times ^{7}C_3$

$=1\times(-35)+(12\times21)+60 \times (-7) + (160 \times 1)$

$= (-35 +252-420+160)$

$=-43$

$\textbf{Therefore, coefficient of } \mathbf{ x^3 } \textbf{is equal to -43}$