Question

Factorise (r^3/8)-(s^3/343)-(t^3/216)-(1/28rst)

Answer 1

Answer:

Step-by-step explanation:

Hi take this one

Answer 2

Concept

Greek mathematicians were the first to think about factorization in the context of integers. They established the arithmetic basic theorem, which states that any positive integer can be factored into a product of prime numbers, but that no prime number can be factored into an integer larger than one. Furthermore, this factorization is exclusive up to the factor order. Even though integer factorization is somewhat of the opposite of multiplication from an algorithmic perspective, the RSA cryptosystem uses this fact to construct public-key cryptography.

Given

$(r^3/8)-(s^3/343)-(t^3/216)-(1/28rst)$

To Find

We have to find the factorized form of $(r^3/8)-(s^3/343)-(t^3/216)-(1/28rst)$.

Solution

According to the problem,

$(r^3/8)-(s^3/343)-(t^3/216)-(1/28rst)$

$=(\frac{r}{2}) ^{3} -(\frac{s}{7}) ^{3} -(\frac{t}{6}) ^{3} -3*\frac{r}{3} \frac{s}{7}\frac{t}{6}$

$=(\frac{r}{2}-\frac{s}{7} -\frac{t}{6} )*((\frac{r}{2}) ^{2} -(\frac{s}{7} ^{2} )-(\frac{t}{6}) ^{2} +\frac{rs}{14} -\frac{st}{42} +\frac{rt}{12} )$

Hence, the factorized for of $(r^3/8)-(s^3/343)-(t^3/216)-(1/28rst)$ $=(\frac{r}{2}-\frac{s}{7} -\frac{t}{6} )*((\frac{r}{2}) ^{2} -(\frac{s}{7} ^{2} )-(\frac{t}{6}) ^{2} +\frac{rs}{14} -\frac{st}{42} +\frac{rt}{12} )$.

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