Question

solve it (2a+b)^2-(2a-b)^3​

Answer 1

$\underbrace{ \boxed{ \tt Solution ☃}}$

$\tt i){(2a + b )}^{2} - {(2a - b)}^{3} \\ \tt using \: eq. \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \tt and \\ \tt {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ \downarrow \downarrow \downarrow \downarrow \\\tt (4 {a}^{2} + 4ab + {b}^{2} ) \times ( {8a}^{3} - 12 {a}^{2} b + 6a {b}^{2} - {b}^{3} ) \\ \tt multiply \: parenthless \\ \downarrow \downarrow \downarrow \downarrow \\ \tt {32a}^{5} - 48 {a}^{4} b + 24 {a}^{3} {b}^{2} - 4 {a}^{2} {b}^{3} + 32 {a}^{4} b - 48 {a}^{3} {b}^{2} + 24 {a}^{2} b^{3} - 4a {b}^{4} + 8 {a}^{3} {b}^{2} - 12 {a}^{2} {b}^{3} + 6a {b}^{4} - {b}^{5} \\ \tt calculate \: like \: terms. \\ \downarrow \downarrow \downarrow \downarrow \\ \tt 32 {a}^{5} - 16 {a}^{4} b - 16 {a}^{3} {b}^{2} + 8 {a}^{2} {b}^{3} + 2a {b}^{4} - {b}^{5} \\ \red { \tt answer} \\ \boxed{ \tt 32 {a}^{5} - 16 {a}^{4} b - 16 {a}^{3} {b}^{2} + 8 {a}^{2} {b}^{3} + 2a {b}^{4} - {b}^{5}}$

Answer 2

$\begin{gathered}\tt i){(2a + b )}^{2} - {(2a - b)}^{3} \\ \tt using \: eq. \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \tt and \\ \tt {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ \downarrow \downarrow \downarrow \downarrow \\\tt (4 {a}^{2} + 4ab + {b}^{2} ) \times ( {8a}^{3} - 12 {a}^{2} b + 6a {b}^{2} - {b}^{3} ) \\ \tt multiply \: parenthless \\ \downarrow \downarrow \downarrow \downarrow \\ \tt {32a}^{5} - 48 {a}^{4} b + 24 {a}^{3} {b}^{2} - 4 {a}^{2} {b}^{3} + 32 {a}^{4} b - 48 {a}^{3} {b}^{2} + 24 {a}^{2} b^{3} - 4a {b}^{4} + 8 {a}^{3} {b}^{2} - 12 {a}^{2} {b}^{3} + 6a {b}^{4} - {b}^{5} \\ \tt calculate \: like \: terms. \\ \downarrow \downarrow \downarrow \downarrow \\ \tt 32 {a}^{5} - 16 {a}^{4} b - 16 {a}^{3} {b}^{2} + 8 {a}^{2} {b}^{3} + 2a {b}^{4} - {b}^{5} \\ \red { \tt answer} \\ \boxed{ \tt 32 {a}^{5} - 16 {a}^{4} b - 16 {a}^{3} {b}^{2} + 8 {a}^{2} {b}^{3} + 2a {b}^{4} - {b}^{5}}\end{gathered}$