Finding the Coefficient of x in Binomial Expansion

The question asks for the coefficient of the term containing 'x' in the expansion of the expression \((x - 3y)^3\). This is a binomial expansion problem. We can use the binomial theorem or the formula for the cube of a binomial, which is \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\).

Applying the Binomial Expansion Formula

In our expression, \((x - 3y)^3\), we can identify \(a = x\) and \(b = -3y\). Now, let's substitute these values into the formula:

Expanding Each Term

Let's calculate each term separately:

• The first term is \((x)^3 = x^3\).
• The second term is \(3(x)^2(-3y) = 3x^2(-3y) = -9x^2y\).
• The third term is \(3(x)(-3y)^2 = 3x(-3)^2(y)^2 = 3x(9y^2) = 27xy^2\).
• The fourth term is \((-3y)^3 = (-3)^3(y)^3 = -27y^3\).

Combining the Terms

Putting all the terms together, the full expansion of \((x - 3y)^3\) is:

Identifying the Coefficient of x

We are looking for the coefficient of the term containing 'x'. In the expanded form, the terms are \(x^3\), \(-9x^2y\), \(27xy^2\), and \(-27y^3\).

The term containing 'x' raised to the power of 1 (which is what 'the coefficient of x' usually refers to unless specified otherwise) is \(27xy^2\).

The coefficient of 'x' in the term \(27xy^2\) is the part that multiplies 'x', which is \(27y^2\).

Comparing with Options

Let's compare our result with the given options:

1. \(3y^2\)
2. \(27y^2\)
3. \(-3y^2\)
4. \(-27y^2\)

Our calculated coefficient of x, which is \(27y^2\), matches the second option.

Therefore, the coefficient of x in \((x - 3y)^3\) is \(27y^2\).

Revision Table: Key Steps for Binomial Expansion

Additional Information on Binomial Coefficients

The binomial theorem provides a systematic way to expand expressions of the form \((a+b)^n\) for any positive integer \(n\).

The general formula is:

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient, also read as "n choose k".

For \((x - 3y)^3\), \(n=3\), \(a=x\), and \(b=-3y\). The terms are generated for \(k=0, 1, 2, 3\):

• \(k=0\): \(\binom{3}{0} x^{3-0} (-3y)^0 = 1 \cdot x^3 \cdot 1 = x^3\)
• \(k=1\): \(\binom{3}{1} x^{3-1} (-3y)^1 = 3 \cdot x^2 \cdot (-3y) = -9x^2y\)
• \(k=2\): \(\binom{3}{2} x^{3-2} (-3y)^2 = 3 \cdot x^1 \cdot (9y^2) = 27xy^2\)
• \(k=3\): \(\binom{3}{3} x^{3-3} (-3y)^3 = 1 \cdot x^0 \cdot (-27y^3) = -27y^3\)

The expansion is \(x^3 - 9x^2y + 27xy^2 - 27y^3\), confirming our previous calculation. The term with \(x^1\) (i.e., \(x\)) is \(27xy^2\), and its coefficient is \(27y^2\).