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Also, can you show us what you have tried so our helpers know where you are stuck and can best help?

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[tex]\text{If }b\text{ is the mean proportion between }a\text{ and }c,[/tex]

[tex]\text{show that: }\:abc(a+b+c)^3 \:=\: (ab+bc+ca)^3[/tex]

We have: .[tex]\frac{a}{b} \,=\,\frac{b}{c} \quad\Rightarrow\quad b^2 \,=\,ac[/tex]

The right side is:

. . [tex](ab+bc + ac)^3 \;=\; (ab + bc + b^2)^3[/tex]

. . [tex]=\;\big[b(a+c +b)\big]^3 \;=\; b^3(a+b+c)^3 [/tex]

. . [tex]=\;b\!\cdot\!b^2(a+b+c)^3 \;=\;b\!\cdot\!ac(a+b+c)^3[/tex]

. . [tex]=\;abc(a+b+c)^3[/tex]

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[tex]\text{If }a,b,c,d \text{ are in continued proportion,}[/tex]

[tex]\text{prove that: }\;\frac{a}{d} \:=\:\frac{a^3 +b^3 +c^3}{b^3 + c^3 +d^3}[/tex]

I assume that

So we have: .[tex]\begin{Bmatrix}a &=& a \\ b &=& ar \\ c &=& ar^2 \\ d &=& ar^3 \end{Bmatrix}[/tex] .where [tex]r[/tex] is the common ratio.

The left side is: .[tex]\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}[/tex]

The right side is: .[tex]\frac{a^3+b^3 + c^3}{b^3+c^3+d^3} \:=\:\frac{a^3 + (ar)^3 + (ar^2)^3}{(ar)^3 + (ar^2)^3 + (ar^3)^3} [/tex]

. . . . [tex]=\;\frac{a^3 + a^3r^3 + a^3r^6}{a^3r^3+a^3r^6+a^3r^9} \;=\; \frac{a^3(1+r^3+r^6)}{a^3r^3(1+r^3+r^6)}[/tex]

. . . . [tex]=\;\frac{1}{r^3}[/tex]

Q.E.D.

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thanks a lot ...Hello again, kuheli!

I assume thatcontinued proportionmeansgeometric sequence.

So we have: .[tex]\begin{Bmatrix}a &=& a \\ b &=& ar \\ c &=& ar^2 \\ d &=& ar^3 \end{Bmatrix}[/tex] .where [tex]r[/tex] is the common ratio.

The left side is: .[tex]\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}[/tex]

The right side is: .[tex]\frac{a^3+b^3 + c^3}{b^3+c^3+d^3} \:=\:\frac{a^3 + (ar)^3 + (ar^2)^3}{(ar)^3 + (ar^2)^3 + (ar^3)^3} [/tex]

. . . . [tex]=\;\frac{a^3 + a^3r^3 + a^3r^6}{a^3r^3+a^3r^6+a^3r^9} \;=\; \frac{a^3(1+r^3+r^6)}{a^3r^3(1+r^3+r^6)}[/tex]

. . . . [tex]=\;\frac{1}{r^3}[/tex]

Q.E.D.

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I have moved your next question to its own thread here:thanks a lot ...

http://mathhelpboards.com/pre-algebra-algebra-2/continued-proportion-problem-7531.html

It is better to create create new threads for new questions. You are more likely to get prompt help this way and our threads do not become long strings of successive problems which are more difficult to follow.