Expanding the given expression and substituting the given values of [tex]\dfrac{p}{q}[/tex] with [tex]\dfrac{q}{r}[/tex] proves that the given equation
Correct response:
[tex]The \ expression \ p^3 + q^3 + r^3 \ is \ equal \ to \ \left(\dfrac{1}{p^3} + \dfrac{1}{q^3} +\dfrac{1}{r^3} \right) \cdot p^2 \cdot q^2 \cdot r^2 \ by \ subtituting[/tex]
[tex]\dfrac{p}{q} = \dfrac{q}{r}[/tex]
Method used to prove that the expression are equal
The given relation is;
[tex]\dfrac{p}{q} = \mathbf{\dfrac{q}{r}}[/tex]
The given equation is presented as follows;
[tex]p^3 + q^3 + r^3 = \mathbf{\left(\dfrac{1}{p^3} + \dfrac{1}{q^3} + \dfrac{1}{r^3} \right) \cdot p^2 \cdot q^2 \cdot r^2}[/tex]
Expanding the right hand side gives;
[tex]\dfrac{p^2 \cdot q^2 \cdot r^2}{p^3} + \dfrac{p^2 \cdot q^2 \cdot r^2}{q^3} + \dfrac{p^2 \cdot q^2 \cdot r^2}{r^3} = \mathbf{ \dfrac{q^2 \cdot r^2}{p} + \dfrac{p^2 \cdot r^2}{q} + \dfrac{p^2 \cdot q^2 }{r}}[/tex]
[tex]\dfrac{q^2 \cdot r^2}{p} + \dfrac{p^2 \cdot r^2}{q} + \dfrac{p^2 \cdot q^2 }{r} = \mathbf{ \dfrac{q}{p} \cdot q \cdot r^2 + \dfrac{p}{q} \cdot p \cdot r^2+\dfrac{q}{r} \cdot p^2 \cdot q }[/tex]
[tex]\dfrac{q}{p} \cdot q \cdot r^2 + \dfrac{p}{q} \cdot p \cdot r^2+\dfrac{q}{r} \cdot p^2 \cdot q } = \dfrac{r}{q} \cdot q \cdot r^2 + \dfrac{q}{r} \cdot p \cdot r^2+\dfrac{p}{q} \cdot p^2 \cdot q } = \mathbf{ r^3 + q \cdot p \cdot r + p^3}[/tex]
From the given relation, we have;
p·r = q²
Therefore;
q·p·r = q × q² = q³
Which gives;
r³ + q·p·r + p³ = r³ + q³ + p³
Which gives;
[tex]\left(\dfrac{1}{p^3} + \dfrac{1}{q^3} + \dfrac{1}{r^3} \right) \cdot p^2 \cdot q^2 \cdot r^2 = p^3 + q^3 + r^3[/tex]
By symmetric property, therefore;
- [tex]\underline{p^3 + q^3 + r^3 = \left(\dfrac{1}{p^3} + \dfrac{1}{q^3} +\dfrac{1}{r^3} \right) \cdot p^2 \cdot q^2 \cdot r^2}[/tex]
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