Math, asked by shimranaktar6, 9 months ago

if p,q,r in A.P then p^3+r^3-8q^3=?

Answers

Answered by yashaswini3679

Answer:

-6pqr

Step-by-step explanation:

Given,

p, q, r are in AP.

So, their common difference will be equal.

i.e, q - p = r - q

2q = r + p

q = (r + p)/2

Substitute this in given equation.

p $^3$ + r $^3$ - 8q $^3$

p $^3$ + r $^3$ - 8(r + p /2) $^3$

p $^3$ + r $^3$ - r $^3$ - p $^3$ - 3pr(r+p)

-3pr(p + r)

-3pr(2q) ...... [p+r = 2q]

-6pqr

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Answered by Sankalp050

Step-by-step explanation:

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$

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if p,q,r in A.P then p^3+r^3-8q^3=?​

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if p,q,r in A.P then p^3+r^3-8q^3=?