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Simplify $\left(a^{2\left(a+1\right)}\right)^3$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2\left(a+1\right)$ and $n$ equals $3$
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$15a^{6\left(a+1\right)}-5a\left(\left(a+1\right)^3\right)^2$
Learn how to solve special products problems step by step online. Expand the expression 15a^(2(a+1))^3-5a(a+1)^3^2. Simplify \left(a^{2\left(a+1\right)}\right)^3 using the power of a power property: \left(a^m\right)^n=a^{m\cdot n}. In the expression, m equals 2\left(a+1\right) and n equals 3. Simplify \left(\left(a+1\right)^3\right)^2 using the power of a power property: \left(a^m\right)^n=a^{m\cdot n}. In the expression, m equals 3 and n equals 2. Multiply the single term 6 by each term of the polynomial \left(a+1\right). We can expand the expression \left(a+1\right)^{6} using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer n. The formula is as follows: \displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n. The number of terms resulting from the expansion always equals n + 1. The coefficients \left(\begin{matrix}n\\k\end{matrix}\right) are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of a decreases, from n to 0, while the exponent of b increases, from 0 to n. If one of the binomial terms is negative, the positive and negative signs alternate..