Find the Product (x-6)^3

The problem asks you to calculate the result of raising the binomial expression (x-6) to the third power. This involves multiplying the expression by itself three times. The operation is an exponentiation of a binomial, and the result will be a polynomial of degree 3. The task involves using algebraic methods, like the binomial theorem or direct multiplication and combining like terms to find the expanded form of the given expression.

${((x - 6))}^{3}$

Answer

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Solution:

Step 1:

Apply the Binomial Theorem to expand $(x - 6)^{3}$ as $x^{3} + 3 x^{2} (- 6) + 3 x (- 6)^{2} + (- 6)^{3}$ .

Step 2:

Simplify each term in the expansion.

Step 2.1:

Multiply $- 6$ by $3 x^{2}$ to get $x^{3} - 18 x^{2} + 3 x (- 6)^{2} + (- 6)^{3}$ .

Step 2.2:

Square $- 6$ to get $x^{3} - 18 x^{2} + 3 x (36) + (- 6)^{3}$ .

Step 2.3:

Multiply $36$ by $3 x$ to get $x^{3} - 18 x^{2} + 108 x + (- 6)^{3}$ .

Step 2.4:

Cube $- 6$ to get the final expanded form $x^{3} - 18 x^{2} + 108 x - 216$ .

Knowledge Notes:

The problem involves expanding a binomial expression raised to a power using the Binomial Theorem. The Binomial Theorem provides a formula for expanding expressions of the form $(a + b)^{n}$ where $n$ is a non-negative integer. The theorem states that:

$(a + b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k}$

Where $(\binom{n}{k})$ is the binomial coefficient, calculated as:

$(\binom{n}{k}) = \frac{n!}{k! (n - k)!}$

For the given problem, the binomial expression is $(x - 6)^{3}$ , which means $a = x$ , $b = - 6$ , and $n = 3$ . The expansion involves terms with powers of $x$ decreasing from 3 to 0 and powers of $- 6$ increasing from 0 to 3.

The steps in the solution involve applying the Binomial Theorem and then simplifying each term by performing arithmetic operations such as multiplication and exponentiation. The final result is the expanded form of the given binomial expression.