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# How do you use pascal’s triangle to expand ${(2x - 3y)^3}?$

Last updated date: 27th Feb 2024
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Hint: As we know that we have to apply the pascal’s triangle to expand the binomial. We know that it is an infinite equilateral triangle which consists of sequence of numbers. It starts with $1$ . The second row consists of the sum of two numbers above it, Similarly we can find out the values of the next rows. It creates a pattern and it is shown below:
$1 \\ 1 - 1 \\ 1 - 2 - 1 \\ 1 - 3 - 3 - 1 \\ \\$

As we know that the main application of this triangle is to solve a binomial function. If the binomial equation is ${(a + b)^n}$ , then the expansion is
${C_1}{a^n}{b^0} + {C_2}{a^{n - 1}}{b^1} + ... + {C_n}{a^0}{b^n}$ .
Since we have $n = 3$ , and from the above triangle we can see that the third order term is $1,3,3,1$ .
For this we have the formula,
${(a - b)^3} = 1{a^3}{b^0} - 3{a^2}{b^1} + 3{a^1}{b^2} - 1{a^0}{b^3}$ .
By applying the above formula we can write
${(2x - 3y)^3} = 1{(2x)^3}{(3y)^0} - 3{(2x)^2}{(3y)^1} + 3{(2x)^1}{(3y)^2} - 1{(2x)^0}{(3y)^3}$ .
On further simplifying we have,
${(2x)^3} - 3{(2x)^2}{(3y)^{}} + 3{(2x)^1}{(3y)^2} - {(3y)^3}$ .
It gives us the expression:
$8{x^3} - 3 \times 4{x^2}{(3y)^{}} + 6x \times 9{y^2} - 27{y^3}$ .
Hence the required value is $8{x^3} - 36{x^2}y + 54x{y^2} - 27{y^3}$ .
So, the correct answer is “ $8{x^3} - 36{x^2}y + 54x{y^2} - 27{y^3}$ ”.

Note: We should note that pascal’s triangle is helpful only when the value of $n$ is small in the equation ${(a + b)^n}$ . If the value is large then it is very tedious to draw the triangle until we reach $n$ . The formula that we used above because the question has a cube of the difference. It there is cube of the sums then the formula that we use is ${(a + b)^3} = 1{a^3}{b^0} + 3{a^2}{b^1} + 3{a^1}{b^2} + 1{a^0}{b^3}$ .