- Thread starter Saumyojit
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It needs to be minimised to a/d .What exactly is it that needs to be solved? a, b, c and d can be solved using only the proportions, and the other expression isn't an equation.

How can I bring it without using k .

most of the people would solve it using a/b = b/c=c/d = kI don't see a k.......

a=bk.....

I dont want to use k

i want to approach it like this

a/b = b/c =>

ac=b^2

b/c=c/d => bd =c^2

see i know thatI see......I was taught that formula differently.

So you want a compressed answer in variable terms?

ac=b^2

b/c=c/d => bd =c^2

how will i substitute

ac=b^2

bd =c^2 in (b^3 + c^3 + d^3)/(a^3+b^3+c^3) ?

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Why not use the k method, which seems simplest?IF a/b = b/c=c/d then solve (b^3 + c^3 + d^3)/(a^3+b^3+c^3)

I dont want an approach using k.

I have started like this :

a/b = b/c =>

ac=b^2

b/c=c/d => bd =c^2

How will i minimize (b^3 + c^3 + d^3)/(a^3+b^3+c^3) ???

I think you mean, not "minimize" (which means, find the lowest value of), but "simplify".

One way is to express c and d in terms of a and b: c = b^2/a, d = c^2/b = b^3/a^2. Make those substitutions, and simplify.see i know that

ac=b^2

b/c=c/d => bd =c^2

how will i substitute

ac=b^2

bd =c^2 in (b^3 + c^3 + d^3)/(a^3+b^3+c^3) ?

Using a/b = b/c = c/d = k (or, as I prefer, b/a = c/b = d/c = k) is a lot less complicated.

I have substituted c = b^2/a, d = c^2/b = b^3/a^2 .Why not use the k method, which seems simplest?

I think you mean, not "minimize" (which means, find the lowest value of), but "simplify".

One way is to express c and d in terms of a and b: c = b^2/a, d = c^2/b = b^3/a^2. Make those substitutions, and simplify.

Using a/b = b/c = c/d = k (or, as I prefer, b/a = c/b = d/c = k) is a lot less complicated.

but i am stuck after 2 steps and i cannot further minimize it .

Please show me the whole steps using this method : c = b^2/a, d = c^2/b = b^3/a^2

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The idea is to express everything in terms of

(b^3 + c^3 + d^3)/(a^3+b^3+c^3) = {b^3+b^6/a^3 +b^9/a^6 }/{a^3+b^3+b^6/a^3)}

=> {b^3 *a^6 +b^3*a^3+b^9}/a^6 * a^3 / {a^6+a^3 * b^3 + b^6} =>( b^3 *a^6 +b^3*a^3+b^9)/(a^6+a^3 * b^3+ b^6) * 1/a^3

after doing lcm

=>b^3 (a^6 +a^3+b^6)/a^3(a^6+a^3 * b^3+ b^6) then what??

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{b^3+b^6/a^3 +b^9/a^6 } = {b^3 *a^6 +b^3*a^3+b^9}/a^6 ?

A major part of doing mathematics well is to learn to check your work and catch errors.

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Did you REVIEW your work and are you certain thatUSING UR METHOD :

(b^3 + c^3 + d^3)/(a^3+b^3+c^3) = {b^3+b^6/a^3 +b^9/a^6 }/{a^3+b^3+b^6/a^3)}

=> {b^3 *a^6 +b^3*a^3+b^9}/a^6 * a^3 / {a^6+a^3 * b^3 + b^6} =>( b^3 *a^6 +b^3*a^3+b^9)/(a^6+a^3 * b^3+ b^6) * 1/a^3

after doing lcm

=>b^3 (a^6 +a^3+b^6)/a^3(a^6+a^3 * b^3+ b^6)then what??

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it is compelted .

dr peterson please have a look in my new post (https://www.freemathhelp.com/forum/threads/what-annual-payment-will-discharge-a-debt-of-rs-580-due-in-5-years-the-rate-being-8-per-si-annum.120207/#post-480913)