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Question Number 122208 by harckinwunmy last updated on 14/Nov/20

$$\mathrm{how}\mathrm{do}\mathrm{i}\mathrm{expand}$$$${(\mathrm{p}+\mathrm{q}+\mathrm{r})}^3?$$$$\mathrm{i}\mathrm{need}\mathrm{general}\mathrm{principle}.$$

Commented by liberty last updated on 14/Nov/20

$$\Rightarrow {(\mathrm{p}+\mathrm{q}+\mathrm{r})}^3=\underset{\mathrm{k}=0}{\sum ^3}\left(\begin{array}{c}3\\ \mathrm{k}\end{array}\right){\mathrm{p}}^{3-\mathrm{k}}.{(\mathrm{q}+\mathrm{r})}^{\mathrm{k}}$$$$=\underset{\mathrm{k}=0}{\sum ^3}\left(\begin{array}{c}3\\ \mathrm{k}\end{array}\right){\mathrm{p}}^{3-\mathrm{k}}[\underset{\ell =0}{\sum ^{\mathrm{k}}}\left(\begin{array}{c}\mathrm{k}\\ \ell \end{array}\right){\mathrm{q}}^{\mathrm{k}-\ell }.{\mathrm{r}}^{\ell }]$$

Commented by liberty last updated on 15/Nov/20

$$\mathrm{k}=0\Rightarrow {\mathrm{p}}^3$$$$\mathrm{k}=1\Rightarrow {3\mathrm{p}}^2(\underset{\ell =0}{\sum ^1}\left(\begin{array}{c}1\\ \ell \end{array}\right){\mathrm{q}}^{1-\ell }.{\mathrm{r}}^{\ell })={3\mathrm{p}}^2(\mathrm{q}+\mathrm{r})$$$$={3\mathrm{p}}^2\mathrm{q}+{3\mathrm{p}}^2\mathrm{r}$$$$\mathrm{k}=2\Rightarrow 3\mathrm{p}(\underset{\ell =0}{\sum ^2}\left(\begin{array}{c}2\\ \ell \end{array}\right){\mathrm{q}}^{2-\ell }.{\mathrm{r}}^{\ell })=3\mathrm{p}({\mathrm{q}}^2+2\mathrm{qr}+{\mathrm{r}}^2)$$$$={3\mathrm{pq}}^2+6\mathrm{pqr}+{3\mathrm{pr}}^2$$$$\mathrm{k}=3\Rightarrow 1.(\underset{\ell =0}{\sum ^3}\left(\begin{array}{c}3\\ \ell \end{array}\right){\mathrm{q}}^{3-\ell }.{\mathrm{r}}^{\ell })={\mathrm{q}}^3+{3\mathrm{q}}^2\mathrm{r}+{3\mathrm{qr}}^2+{\mathrm{r}}^3$$$$\mathrm{therefore}\mathrm{we}\mathrm{get}$$$${(\mathrm{p}+\mathrm{q}+\mathrm{r})}^3={\mathrm{p}}^3+{3\mathrm{p}}^2\mathrm{q}+{3\mathrm{p}}^2\mathrm{r}+{3\mathrm{pq}}^2+6\mathrm{pqr}+{3\mathrm{pr}}^2$$$$+{\mathrm{q}}^3+{3\mathrm{q}}^2\mathrm{r}+{3\mathrm{qr}}^2+{\mathrm{r}}^3.$$

Answered by mathmax by abdo last updated on 15/Nov/20

$${(\mathrm{p}+\mathrm{q}+\mathrm{r})}^3={(\mathrm{p}+\mathrm{q}+\mathrm{r})}^2(\mathrm{p}+\mathrm{q}+\mathrm{r})$$$$=({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2+2\mathrm{pq}+2\mathrm{pr}+2\mathrm{qr})(\mathrm{p}+\mathrm{q}+\mathrm{r})$$$$={\mathrm{p}}^3+{\mathrm{p}}^2\mathrm{q}+{\mathrm{p}}^2\mathrm{r}+{\mathrm{q}}^2\mathrm{p}+{\mathrm{q}}^3+{\mathrm{q}}^2\mathrm{r}+{\mathrm{r}}^2\mathrm{p}+{\mathrm{r}}^2\mathrm{q}+{\mathrm{r}}^3++{2\mathrm{p}}^2\mathrm{q}+{2\mathrm{pq}}^2+2\mathrm{pqr}$$$$+{2\mathrm{p}}^2\mathrm{r}+2\mathrm{prq}+{2\mathrm{pr}}^2+2\mathrm{qrp}+{2\mathrm{q}}^2\mathrm{r}+{2\mathrm{qr}}^2=....$$

Answered by $@y@m last updated on 15/Nov/20

$$Step\left(i\right)Putp+q=a$$$$Step\left(ii\right)Expand{(a+r)}^3$$$$Step\left(iii\right)Puta=p+qandsimplify.$$

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