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Using the identity $$\prod\limits_{n=1}^{\infty}\left(1-x^{2n}\right)\left(1+yx^{2n-1}\right)\left(1+y^{-1}x^{2n-1}\right)=\sum\limits_{n\in\mathbb Z}y^nx^{n^2},$$ show that $$ \left[\prod\limits_{n=1}^{\infty}(1-a^n)\right]^3=\sum\limits_{n=0}^{\infty}(-1)^n(2n+1)a^{\frac{n(n+1)}{2}}.$$

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    is the identity correct? what if $x=y=1$? – 2017-01-10
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    @ArnaldoNascimento it's valid for $\lvert x\rvert < 1$ and $y \neq 0$. And Li Li, what substitutions have you tried? – 2017-01-10
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    I would start with $x=\sqrt{a}$ and $y=-\sqrt{a}$ – 2017-01-10
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    @kobe The same with the try of Arnaldo Nascimento. But I don't know how to continue. – 2017-01-10

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The first identity in question is more popularly known as Jacobi Triple Product identity and the identity to be proved is also known by the name of Jacobi's identity. Getting the second identity from the first is not direct, but at the same time not too difficult.

Let's put $y = -xt$ where $t$ is real and $t \neq 1$ to get $$\prod_{n = 1}^{\infty}(1 - x^{2n})(1 - x^{2n}t)(1 - x^{2n - 2}t^{-1}) = \sum_{n = -\infty}^{\infty}(-1)^{n}t^{n}x^{n(n + 1)}$$ or $$(t - 1)\prod_{n = 1}^{\infty}(1 - x^{2n})(1 - x^{2n}t)(1 - x^{2n}t^{-1}) = \sum_{n = -\infty}^{\infty}(-1)^{n}t^{n + 1}x^{n(n + 1)}$$ Now we can pair the terms corresponding to indices $n$ and $-(n + 1)$ in the sum on right to get $$(t - 1)\prod_{n = 1}^{\infty}(1 - x^{2n})(1 - x^{2n}t)(1 - x^{2n}t^{-1}) = \sum_{n = 0 }^{\infty}(-1)^{n}x^{n(n + 1)}\{t^{n + 1} - t^{-n}\}$$ or $$\prod_{n = 1}^{\infty}(1 - x^{2n})(1 - x^{2n}t)(1 - x^{2n}t^{-1}) = \sum_{n = 0 }^{\infty}(-1)^{n}x^{n(n + 1)}\cdot\frac{t^{n + 1} - t^{-n}}{t - 1}$$ Letting $t \to 1$ we get $$\prod_{n = 1}^{\infty}(1 - x^{2n})^{3} = \sum_{n = 0}^{\infty}(-1)^{n}(2n + 1)x^{n(n + 1)}$$ Replacing $x$ by $\sqrt{a}$ we get the desired identity in question.