$${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$$

$$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$$

$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$

$$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$$

$$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$$

$$ = 2025 - 1600 = 425$$

Let $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$

Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$

Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$

$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$

Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$ to $$\infty = 100$$

$$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$$

$$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$$

From $$(i),$$ $${a^2} = 400{\left( {1 - r} \right)^2};$$

From $$(ii),$$ we get $$100\left( {1 - r} \right)\left( {1 + r} \right)$$

$$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$$

$$ \Rightarrow 1 + r = 4 - 4r$$

$$ \Rightarrow 5r = 3$$

$$ \Rightarrow r = 3/5.$$

$$a{r^4} = 2$$

$$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$$

$$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$$

The product is $$p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$$

$$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$

Now let

$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$

$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$$

Subtracting $$(2)$$ from $$(1)$$

$$ \Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty $$

or $${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$$

$$\therefore$$ $$P = {2^S} = 2$$