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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 .

Commented by liberty last updated on 14/Nov/20

\Rightarrow {(p + q + r)}^{3} = \underset{k = 0}{\sum^{3}} (\begin{matrix} 3 \\ k \end{matrix}) p^{3 - k} . {(q + r)}^{k}

= \underset{k = 0}{\sum^{3}} (\begin{matrix} 3 \\ k \end{matrix}) p^{3 - k} [\underset{ℓ = 0}{\sum^{k}} (\begin{matrix} k \\ ℓ \end{matrix}) q^{k - ℓ} . r^{ℓ}]

Commented by liberty last updated on 15/Nov/20

k = 0 \Rightarrow p^{3}

k = 1 \Rightarrow {3 p}^{2} (\underset{ℓ = 0}{\sum^{1}} (\begin{matrix} 1 \\ ℓ \end{matrix}) q^{1 - ℓ} . r^{ℓ}) = {3 p}^{2} (q + r)

= {3 p}^{2} q + {3 p}^{2} r

k = 2 \Rightarrow 3 p (\underset{ℓ = 0}{\sum^{2}} (\begin{matrix} 2 \\ ℓ \end{matrix}) q^{2 - ℓ} . r^{ℓ}) = 3 p (q^{2} + 2 qr + r^{2})

= {3 pq}^{2} + 6 pqr + {3 pr}^{2}

k = 3 \Rightarrow 1 . (\underset{ℓ = 0}{\sum^{3}} (\begin{matrix} 3 \\ ℓ \end{matrix}) q^{3 - ℓ} . r^{ℓ}) = q^{3} + {3 q}^{2} r + {3 qr}^{2} + r^{3}

therefore we get

{(p + q + r)}^{3} = p^{3} + {3 p}^{2} q + {3 p}^{2} r + {3 pq}^{2} + 6 pqr + {3 pr}^{2}

+ q^{3} + {3 q}^{2} r + {3 qr}^{2} + r^{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} + 2 pq + 2 pr + 2 qr) (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} + + {2 p}^{2} q + {2 pq}^{2} + 2 pqr

+ {2 p}^{2} r + 2 prq + {2 pr}^{2} + 2 qrp + {2 q}^{2} r + {2 qr}^{2} = \dots .

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

S t e p (i) P u t p + q = a

S t e p (i i) E x p a n d {(a + r)}^{3}

S t e p (i i i) P u t a = p + q a n d s i m p l i f y .

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