# mean proportion problem

#### kuheli

##### New member
if b is the mean proportion between a and c ,show that

abc(a+b+c)^3 = (ab+bc+ca)^3

#### MarkFL

Staff member

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

#### kuheli

##### New member
i am unable to figure out in which direction and how to proceed with this

#### soroban

##### Well-known member
Hello, kuheli!

$$\text{If }b\text{ is the mean proportion between }a\text{ and }c,$$

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

We have: .$$\frac{a}{b} \,=\,\frac{b}{c} \quad\Rightarrow\quad b^2 \,=\,ac$$

The right side is:

. . $$(ab+bc + ac)^3 \;=\; (ab + bc + b^2)^3$$

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

. . $$=\;b\!\cdot\!b^2(a+b+c)^3 \;=\;b\!\cdot\!ac(a+b+c)^3$$

. . $$=\;abc(a+b+c)^3$$

#### kuheli

##### New member
hi ... thanks a lot .
there is another problem with which i am stuck .i tried but its getting more complicated and long without any positive sign.the numerical is:

if a,b,c,d are in continued proportion prove that

a/d=(a^3 +b^3 +c^3)/(b^3 + c^3 +d^3)

#### soroban

##### Well-known member
Hello again, kuheli!

$$\text{If }a,b,c,d \text{ are in continued proportion,}$$

$$\text{prove that: }\;\frac{a}{d} \:=\:\frac{a^3 +b^3 +c^3}{b^3 + c^3 +d^3}$$

I assume that continued proportion means geometric sequence.

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

The left side is: .$$\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}$$

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

. . . . $$=\;\frac{a^3 + a^3r^3 + a^3r^6}{a^3r^3+a^3r^6+a^3r^9} \;=\; \frac{a^3(1+r^3+r^6)}{a^3r^3(1+r^3+r^6)}$$

. . . . $$=\;\frac{1}{r^3}$$

Q.E.D.

#### kuheli

##### New member
Hello again, kuheli!

I assume that continued proportion means geometric sequence.

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

The left side is: .$$\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}$$

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

. . . . $$=\;\frac{a^3 + a^3r^3 + a^3r^6}{a^3r^3+a^3r^6+a^3r^9} \;=\; \frac{a^3(1+r^3+r^6)}{a^3r^3(1+r^3+r^6)}$$

. . . . $$=\;\frac{1}{r^3}$$

Q.E.D.
thanks a lot ...