How do you expand ${(3 a - b)}^{4}$ $(3a-b)^4$ using Pascal’s Triangle?

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How do you expand ${(3 a - b)}^{4}$ $(3a-b)^4$ using Pascal’s Triangle?

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Answer 1 · 2016-02-03T11:16:14

$81 a^{4} - 108 a^{3} b + 54 a^{2} b^{2} - 12 a b^{3} + b^{4}$ $81a^4-108a^3b+54a^2b^2-12ab^3+b^4$

Explanation:

from pascal's triangle we can see,

${(1 + x)}^{0}$ $(1+x)^0$ --------------------- $1$ $1$
${(1 + x)}^{1}$ $(1+x)^1$ --------------------- $1$ $1$ $1$ $1$
${(1 + x)}^{2}$ $(1+x)^2$ -------------------- $1$ $1$ $2$ $2$ $1$ $1$
${(1 + x)}^{3}$ $(1+x)^3$ -------------------- $1$ $1$ $3$ $3$ $3$ $3$ $1$ $1$
${(1 + x)}^{4}$ $(1+x)^4$ -------------------- $1$ $1$ $4$ $4$ $6$ $6$ $4$ $4$ $1$ $1$

so,
${(3 a - b)}^{4} = 1 \cdot {(3 a)}^{4} - 4 \cdot {(3 a)}^{3} b + 6 \cdot {(3 a)}^{2} b^{2} - 4 \cdot 3 a b^{3} + 1 \cdot b^{4}$ $(3a-b)^4=1*(3a)^4-4*(3a)^3b+6*(3a)^2b^2-4*3ab^3+1*b^4$
$= 81 a^{4} - 108 a^{3} b + 54 a^{2} b^{2} - 12 a b^{3} + b^{4}$ $=81a^4-108a^3b+54a^2b^2-12ab^3+b^4$

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How do you expand ${(3 a - b)}^{4}$ $(3a-b)^4$ using Pascal’s Triangle?

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