Question

expansion of (a+3b) power 4

Answer 1

a^4 + 4a^33b + 6a^29b^2 + 4a27b^3 + 81b^4

Answer 2

Heya user !!

Here's the answer you are looking for

(a + 3b)⁴

By Binomial theorem,

${(x + y)}^{n}$
$= nC0\: {x}^{n} {y}^{0} + nC1 \: {x}^{n - 1} {y}^{1} + ...... + nCn \: {x}^{0} {y}^{n}$

Similarly, here,

(a + 3b)⁴

$= 4C0 \: {a}^{4} {(3b)}^{0} + 4C1 \: {a}^{3} {(3b)}^{1} + 4C2 \: {a}^{2} {(3b)}^{2} \\ + 4C3 \: {a}^{1} {(3b)}^{3} + 4C4 \: {a}^{0} {(3b)}^{4} \\ \\ = {a}^{4} + 4 {a}^{3} (3b) + 6 {a}^{2} ( {9b}^{2} ) \\ + 4a( {27b}^{3} ) + 81 {b}^{4} \\ \\ = {a}^{4} + 12 {a}^{3} b + 54 {a}^{2} {b}^{2} + 108a {b}^{3} + 81 {b}^{3}$

▶️OTHER WAY (WITHOUT BINOMIAL THEOREM)◀️

(a + 3b)⁴

= (a + 3b)² × (a + 3b)²

= (a² + 9b² + 6ab) × (a² + 9b² + 6ab)

= a²(a² + 9b² + 6ab) + 9b²(a² + 9b² + 6ab) + 6ab(a² + 9b² + 6ab)

= a⁴ + 9a²b² + 6a³b + 9a²b² + 81b⁴ + 54ab³ + 6a³b + 54ab³ + 36a²b²

REARRANGE

= a⁴ + (6a³b + 6a³b) + (9a²b² + 9a²b² + 36a²b²) + (54ab³ + 54ab³) + 81b⁴

= a⁴ + 12a³b + 54a²b² + 108ab³ + 81b³


★★ HOPE THAT HELPS ☺️ ★★