Question

625r^6-5q^6

Answer 1

Is it to be factorised???

Answer 2

yes im sorry

Answer 3

what about it? this is pretty much simplified already. you could factor this into\[(25r^3-\sqrt{5}q^3)(25r^3+\sqrt{5}q^3)\]

Answer 4

u have the answer just dont know if it is right, i get lost after i find the lcm

Answer 5

i have i mean

Answer 6

I just factored it with the difference of squares. recognize that this is like a^2-b^2 factors into
(a-b)(a+b)

Answer 7

i didnt get that answer

Answer 8

maybe i messed up multiplying

Answer 9

so just take the square root of each term 625r^6 and 5q^6 and put it in this form:
(25r3−5√q3)(25r3+5√q3)
I see nop other way to factor this.

Answer 10

yea thats along the line i got turnigtest

Answer 11

Delta_r = 13904571533203125000000 q^30

Answer 12

r = ((-1)^(1/3) q)/sqrt(5)

Answer 13

\[(25r^3-\sqrt{5}q^3)(25r^3+\sqrt{5}q^3)=625r^6-25\sqrt{5}q^3r^3+25\sqrt{5}q^3r^3-5q^6\]

Answer 14

which gives you your original expression

Answer 15

5(5r^2-q^2)(25r^4+5r^2q^@+q^4

Answer 16

something along that line lol

Answer 17

minus the typos. thanks for the help

Answer 18

don't factor out the 5 first if you can help it. that makes the first coefficient 125 which is no longer a perfect square. Just follow the format:

a^2-b^2
factors into
(a-b)(a+b)

Answer 19

625r^6-5q^6 = 5(125r^6 - q^6) = 5 [(5r^2)^3 - (q^2)^3]
using a^3 - b^3 = (a - b) (a^2 + ab + b^2)
= 5 [(5r^2 - q^2)(25r^4 + 5r^2q^2 + q^4)]
= 5(5r^2 - q^2)(25r^4 + 5r^2q^2 + q^4)

Hope this is clear.....

Answer 20

yea i think i just messed up on the factoring part