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

how do i expand (p+q+r)^3 ? i need general principle.

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

Commented by liberty last updated on 14/Nov/20

⇒(p+q+r)^3 = Σ_(k = 0) ^3 ((3),(k) ) p^(3−k) .(q+r)^k = Σ_(k=0) ^3 ((3),(k) ) p^(3−k) [ Σ_(ℓ=0) ^k ((k),(ℓ) ) q^(k−ℓ) .r^ℓ ]

$$\Rightarrow\left(\mathrm{p}+\mathrm{q}+\mathrm{r}\right)^{\mathrm{3}} \:=\:\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\begin{pmatrix}{\mathrm{3}}\\{\mathrm{k}}\end{pmatrix}\:\mathrm{p}^{\mathrm{3}−\mathrm{k}} .\left(\mathrm{q}+\mathrm{r}\right)^{\mathrm{k}} \\ $$$$\:=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\begin{pmatrix}{\mathrm{3}}\\{\mathrm{k}}\end{pmatrix}\:\mathrm{p}^{\mathrm{3}−\mathrm{k}} \:\left[\:\underset{\ell=\mathrm{0}} {\overset{\mathrm{k}} {\sum}}\begin{pmatrix}{\mathrm{k}}\\{\ell}\end{pmatrix}\:\mathrm{q}^{\mathrm{k}−\ell} .\mathrm{r}^{\ell} \:\right]\: \\ $$

Commented by liberty last updated on 15/Nov/20

k=0⇒ p^3 k=1⇒3p^2 (Σ_(ℓ=0) ^1 ((1),(ℓ) ) q^(1−ℓ) .r^ℓ )=3p^2 (q+r) = 3p^2 q+3p^2 r k=2⇒3p (Σ_(ℓ=0) ^2 ((2),(ℓ) ) q^(2−ℓ) .r^ℓ )=3p(q^2 +2qr+r^2 ) = 3pq^2 +6pqr+3pr^2 k=3⇒1.(Σ_(ℓ=0) ^3 ((3),(ℓ) ) q^(3−ℓ) .r^ℓ )=q^3 +3q^2 r+3qr^2 +r^3 therefore we get (p+q+r)^3 = p^3 +3p^2 q+3p^2 r+3pq^2 +6pqr+3pr^2 + q^3 +3q^2 r+3qr^2 +r^3 .

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

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

(p+q+r)^3 =(p+q+r)^2 (p+q+r) =(p^2 +q^2 +r^2 +2pq +2pr +2qr)(p+q+r) =p^3 +p^2 q+p^2 r +q^2 p+q^3 +q^2 r +r^2 p+r^2 q +r^3 ++2p^2 q+2pq^2 +2pqr +2p^2 r +2prq +2pr^2 +2qrp +2q^2 r +2qr^2 =....

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

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

Step (i) Put p+q=a Step (ii) Expand (a+r)^3 Step (iii) Put a= p+q and simplify.

$${Step}\:\left({i}\right)\:{Put}\:\:{p}+{q}={a} \\ $$$${Step}\:\left({ii}\right)\:{Expand}\:\left({a}+{r}\right)^{\mathrm{3}} \\ $$$${Step}\:\left({iii}\right)\:{Put}\:{a}=\:{p}+{q}\:{and}\:{simplify}. \\ $$

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