Question

Ratio and proportion.

Answer 1

@gudden

Answer 2

(I'm asking this on their behalf due to some problems on their end.)

Answer 3

If\[\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\]then\[\frac{ x ^{3} + a ^{3} }{ x ^{2} + a ^{2}} + \frac{ y ^{3} + b ^{3}} {a ^{2} + b ^{2} } + \frac{ z ^{3} + c ^{3}}{ z ^{2} + c ^{2} } = \frac{ (x+y+z)^{3} + (a+b+c)^{3} }{ (x+y+z)^{2} +(a +b+c)^{2} }\]

Answer 4

Let's say that\[\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c} =k\]then\[x=ka\]\[y=kb\]\[z=kc\]

Answer 5

yeah and then replacing these values of x,y and z in the actual equation / LHS we can get the RHS :p
Thanks for the help...:)
I needed the starting step badly..! ( was sorta, puzzled with what to start with :) )

Answer 6

We probably need a little bit of manipulation on this\[\dfrac{(ka + kb + kc)^3 + (a+b+c)^3}{(ka + kb + kc)^2 + (a+b+c)^2}\]\[=\dfrac{ (k^3+1) \left(a + b + c\right)^3}{(k^2+1)(a + b + c)^2}\]\[=\dfrac{k^3+1}{k^2+1}(a+b+c)\]You'll get the same expression for the LHS.

Answer 7

Yup, that's right.

Answer 8

Where are you doing these problems from?

Answer 9

Do you want them ??
They are actually in my booklet!

Answer 10

Most of them are calculus based, but the last question at the page is of the old concepts...( those done in 9th and 10th)

Answer 11

*at every page

Answer 12

Which booklet?

Answer 13

Booklet of our Tuition center..!