Question

The HCF and LCM of three expressions are a^3bc^2 and a^5b^3c^4 respectively. if two of them are a^3bc^3 and a^4b^2c^4 then the third is?

Answer 1

sorry what math is this?

Answer 2

if its in a math book then you can look it up on Slader.com

Answer 3

i want to know how to do it

Answer 4

its very tricky

Answer 5

their is is formula for
when \(a\) and \(b\) are two positive integers
then
\(a\times b=HCF\times LCM\)


but not that case if their are three
if \(a\) ,\(b\) and \(c\) are three positive integers then

\(a\times b\times c\neq HCF\times LCM\)

Answer 6

no offense but that doesnt really help

Answer 7

\(\large\tt \begin{align} \color{black}{x\times y\times z=\dfrac{LCM(x,y,z)\times HCF(x,y)\times HCF(y,z)\times HCF(z,x)}{HCF(x,y,z)}}\end{align}\)

Answer 8

is the answer or solution given

Answer 9

to me, it is \(a^5 b^3 c^2\)

Answer 10

if the first number is \(N_1=a^3bc^3\) and the second one is \(N_2=a^4b^2c^4\), then \(N_3=a^mb^nc^l\)

Answer 11

for LCM \(a^5b^3c^4\), it forces m=5, n=3 since the first term and the second term have the exponent of a and b <5

Answer 12

for HCF \(a^3bc^2\), it forces l =2
Hence the third one is \(a^5b^3c^2\)

Answer 13

*the exponent of a for \(N_1,N_2<5\) in LCM, that forces the exponent of a in \(N_3\) is 5

Answer 14

the exponent of b for \(N_1,N_2<3\) in LCM, that forces the exponent of b in \(N_3\) is 3

Answer 15

and the exponent of c relies on HCF which is 2

Answer 16

wait

Answer 17

yes, sir. :)

Answer 18

wolfram is giving the third one
as \(\Large (a,b)\)

http://www.wolframalpha.com/input/?i=%28gcd%28a%5E3bc%5E3%2Ca%5E4b%5E2c%5E4%2Cz%29%3Da%5E3bc%5E2%2C+lcm%28a%5E3bc%5E3%2Ca%5E4b%5E2c%5E4%2Cz%29%3Da%5E5b%5E3c%5E4+%29

Answer 19

I don't know, but if c^4 is the answer on the third one, then the c in gcd is not c^2, it must be c^3

Answer 20

its not \(\huge c^4\) its is giving \(\huge z=ab\)