Question

simplify:-(x/3+y/5)³-(x/3-y/5)³​

Answer 1

Answer:

i) If the fourth verse 16 and seventh verse 128 of a Gu.

3. Remember Kisagu is the (m + n) or medal and (m-n) va term of 00, then find the m and va term of the series

4. The product of the third and eighth terms of a Gu is 243. If its fourth term is 3, find the seventh term.

5. If the n term of the range 1 + 2 + 4 + ... is the same as the n term of the range 256 + 128 + 64 + ... then find the value of n

6. Prove for a particular category, xtar = (ER) -

7. The rank of a Gu 00 is a, 14th, b and 24, then show that b * -afc

(Hint: = AR: b = AR13: c = ARD. C = (ARB) - (ARP)]

Lp

8. If the p and tave terms of a given category are respectively p and p, then prove that the p (v +) term of the series is

9. The third term of a geometric series is the square of the first term, and the fifth term is 64, find the series.

10. Prove in a certain category that the product of the positions of the same distance from the beginning and the end, the first term and s

Is equal to the product.

11. If they, they and those terms of a Gulshray 0 are x.y,: respectively, then show that x-y-

(U.P. diploma

Answer 2

$\huge\underline\mathfrak\pink{Solution:}$

We have,

(x/3+y/5)³-(x/3-y/5)³ = $\sf{(a+b)^3}$-$\sf{(a-b)^3, where}$$\sf\frac{x}{3}$$\sf{=a\:and}$$\sf\frac{x}{3}$$\sf{=b}$

____________________________

$\sf{[a^3+b^3+3ab(a+b)]-[a^3-b^3-3ab(a-b)]}$

$\sf{[a^3+b^3+3a^2b+3ab^2]-[a^3-b^3-3a^2b+3ab^2]}$

$\sf{a^3+b^3+3a^2b+3ab^2-a^3+b^3+3a^2b-3ab^2}$

$\sf{=2b^3+6a^2b}$

=2$\sf\frac{(y)}{(5)}^3$+6$\sf\frac{(x)}{(3)}^2$$\sf\frac{(y)}{(5)}$,

$\bold{\large{\boxed{\sf{\red{Putting\:back\:values\:of\:a=x/3\:and\:b=y/5)}}}}}$

$\huge\sf\frac{2}{125}$y³ + $\huge\sf\frac{2}{15}$x²y

_____________________________

$\bold{\large{\boxed{\sf{\green{Answer:2/5[y]^3/[25]+[x^2y]/[3]}}}}}$