Q. Expand using the binomial theorem \( \left(x^{2}\right)^{6} \).

Q: What is the binomial theorem for (a+b)^n?

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

Q: How do you rewrite (x^2)^6 before expanding?

(x^2)^6=x^{2\cdot 6}=x^{12}. No binomial expansion is needed since it’s a single-term power.

Q: If (1+x^2)^6 were intended, what’s the expanded polynomial?

(1+x^2)^6=1+6x^2+15x^4+20x^6+15x^8+6x^{10}+x^{12}.

Q: Why does (x^2)^6 not require binomial theorem?

Binomial theorem applies to (a+b)^n. Here (x^2)^6 is (a)^n with a=x^2, so simplify directly to x^{12}.

Answer

Use the binomial theorem:

\[
(x^2+0)^6=\sum_{k=0}^{6}\binom{6}{k}(x^2)^k(0)^{6-k}.
\]
All terms vanish except when \(6-k=0\), so \(k=6\). Thus,
\[
(x^2)^6=x^{12}.
\]

Final result: \(\,x^{12}\)

Detailed Explanation

We want to expand using the binomial theorem: \((x^{2})^{6}\).

Step 1: Rewrite the power

Use the exponent rule \((a^{m})^{n} = a^{mn}\). Here \(a = x\), \(m = 2\), and \(n = 6\):

\[
(x^{2})^{6} = x^{2 \cdot 6} = x^{12}
\]

Step 2: Apply the binomial theorem

The binomial theorem expands expressions of the form \((a+b)^n\). But \((x^{2})^{6}\) is not a binomial in the form \((a+b)^n\); it is just a single term being raised to a power.

Final answer

\[
(x^{2})^{6} = x^{12}
\]

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Algebra FAQ

What is the binomial theorem for \((a+b)^n\)?

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

How do you rewrite \((x^2)^6\) before expanding?

\((x^2)^6=x^{2\cdot 6}=x^{12}\). No binomial expansion is needed since it’s a single-term power.

If the problem mistakenly meant \((1+x^2)^6\), what is the expansion form?

\((1+x^2)^6=\sum_{k=0}^{6}\binom{6}{k}1^{6-k}(x^2)^k=\sum_{k=0}^{6}\binom{6}{k}x^{2k}\).

If \((1+x^2)^6\) were intended, what’s the expanded polynomial?

\((1+x^2)^6=1+6x^2+15x^4+20x^6+15x^8+6x^{10}+x^{12}\).

What are the binomial coefficients \(\binom{6}{k}\) used for \(k=0\) to \(6\)?

\(\binom{6}{0}=1,\binom{6}{1}=6,\binom{6}{2}=15,\binom{6}{3}=20,\binom{6}{4}=15,\binom{6}{5}=6,\binom{6}{6}=1\).

Why does \((x^2)^6\) not require binomial theorem?

Binomial theorem applies to \((a+b)^n\). Here \((x^2)^6\) is \((a)^n\) with \(a=x^2\), so simplify directly to \(x^{12}\).

Use binomial theorem for (x^2)^6.
Expand and simplify the terms.

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