Expand Using Pascal's Triangle (2x^3-3y)^3

${(2 x^{3} - 3 y)}^{3}$

Step 1

Pascal\'s Triangle can be displayed as such:

$1$

$1 - 1$

$1 - 2 - 1$

$1 - 3 - 3 - 1$

The triangle can be used to calculate the coefficients of the expansion of ${(a + b)}^{n}$ by taking the exponent $n$ and adding $1$ . The coefficients will correspond with line $n + 1$ of the triangle. For ${(2 x^{3} - 3 y)}^{3}$ , $n = 3$ so the coefficients of the expansion will correspond with line $4$ .

Step 2

The expansion follows the rule ${(a + b)}^{n} = c_{0} a^{n} b^{0} + c_{1} a^{n - 1} b^{1} + c_{n - 1} a^{1} b^{n - 1} + c_{n} a^{0} b^{n}$ . The values of the coefficients, from the triangle, are $1 - 3 - 3 - 1$ .

$1 a^{3} b^{0} + 3 a^{2} b + 3 a b^{2} + 1 a^{0} b^{3}$

Step 3

Substitute the actual values of $a$ $2 x^{3}$ and $b$ $- 3 y$ into the expression.

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

Step 4

Simplify each term.

Multiply ${(2 x^{3})}^{3}$ by $1$ .

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

Apply the product rule to $2 x^{3}$ .

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

Raise $2$ to the power of $3$ .

$8 {(x^{3})}^{3} {(- 3 y)}^{0} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply the exponents in ${(x^{3})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 x^{3 \cdot 3} {(- 3 y)}^{0} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $3$ by $3$ .

$8 x^{9} {(- 3 y)}^{0} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Apply the product rule to $- 3 y$ .

$8 x^{9} ({(- 3)}^{0} y^{0}) + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Rewrite using the commutative property of multiplication.

$8 \cdot {(- 3)}^{0} x^{9} y^{0} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Anything raised to $0$ is $1$ .

$8 \cdot 1 x^{9} y^{0} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $8$ by $1$ .

$8 x^{9} y^{0} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Anything raised to $0$ is $1$ .

$8 x^{9} \cdot 1 + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $8$ by $1$ .

$8 x^{9} + 3 {(2 x^{3})}^{2} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Apply the product rule to $2 x^{3}$ .

$8 x^{9} + 3 (2^{2} {(x^{3})}^{2}) {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Raise $2$ to the power of $2$ .

$8 x^{9} + 3 (4 {(x^{3})}^{2}) {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply the exponents in ${(x^{3})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 x^{9} + 3 (4 x^{3 \cdot 2}) {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $3$ by $2$ .

$8 x^{9} + 3 (4 x^{6}) {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $4$ by $3$ .

$8 x^{9} + 12 x^{6} {(- 3 y)}^{1} + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Simplify.

$8 x^{9} + 12 x^{6} (- 3 y) + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Rewrite using the commutative property of multiplication.

$8 x^{9} + 12 \cdot - 3 x^{6} y + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $12$ by $- 3$ .

$8 x^{9} - 36 x^{6} y + 3 {(2 x^{3})}^{1} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Simplify.

$8 x^{9} - 36 x^{6} y + 3 (2 x^{3}) {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $2$ by $3$ .

$8 x^{9} - 36 x^{6} y + 6 x^{3} {(- 3 y)}^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Apply the product rule to $- 3 y$ .

$8 x^{9} - 36 x^{6} y + 6 x^{3} ({(- 3)}^{2} y^{2}) + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Rewrite using the commutative property of multiplication.

$8 x^{9} - 36 x^{6} y + 6 \cdot {(- 3)}^{2} x^{3} y^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Raise $- 3$ to the power of $2$ .

$8 x^{9} - 36 x^{6} y + 6 \cdot 9 x^{3} y^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply $6$ by $9$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + 1 {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Multiply ${(2 x^{3})}^{0}$ by $1$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + {(2 x^{3})}^{0} {(- 3 y)}^{3}$

Apply the product rule to $2 x^{3}$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + 2^{0} {(x^{3})}^{0} {(- 3 y)}^{3}$

Anything raised to $0$ is $1$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + 1 {(x^{3})}^{0} {(- 3 y)}^{3}$

Multiply ${(x^{3})}^{0}$ by $1$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + {(x^{3})}^{0} {(- 3 y)}^{3}$

Multiply the exponents in ${(x^{3})}^{0}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + x^{3 \cdot 0} {(- 3 y)}^{3}$

Multiply $3$ by $0$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + x^{0} {(- 3 y)}^{3}$

Anything raised to $0$ is $1$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + 1 {(- 3 y)}^{3}$

Multiply ${(- 3 y)}^{3}$ by $1$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + {(- 3 y)}^{3}$

Apply the product rule to $- 3 y$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} + {(- 3)}^{3} y^{3}$

Raise $- 3$ to the power of $3$ .

$8 x^{9} - 36 x^{6} y + 54 x^{3} y^{2} - 27 y^{3}$

Expand Using Pascal's Triangle (2x^3-3y)^3

Related Questions

Support LCMGCF.com by disabling your adblocker.