Question

I want answer to the Q-43

Answer 1

Hello mate!!! ____^_^___
Here is the answer. .....

If a+b+c= 0
then a^3+b^3+c^3= 3abc.

HERE, a= r (s-t)
b= s (t-r)
c= t (r-s).
now,
3+b^3+c^3= 3abc.
=> 3rst (s-t) (r-s) (t-r).


hope it helps ☺✌❤

Answer 2

$\huge{\bold{\boxed{\boxed{\mathfrak{\red{Your\:Answer}}}}}}$

$\huge{\bold{\textsf{\pink{We Have To Factorise}}}}$

$\mathfrak{\blue{r^3(s-t)^3+s^3(t-r)^3+t^3(r-s)^3}}$

= {r(s-t)}³+{s(t-r)}³+t(r-s)}³

Now it's in the form $\mathfrak{\red{a^3+b^3+c^3}}$

$\text{SO\:WE\:HAVE}$

$\huge{\bold{\boxed{\boxed{\mathfrak{\red{ a=r(s-t)}}}}}}$

$\huge{\bold{\boxed{\boxed{\mathfrak{\red{ b=s(t-r)}}}}}}$

$\huge{\bold{\boxed{\boxed{\mathfrak{\red{c=t(r-s)}}}}}}$

now $a+b+c=r(s-t)+s(t-r)+t(r-s)$

$=rs-rt+st-sr+tr-ts$

$=rs-sr+st-ts+tr-rt$

$=rs-rs+st-st+tr-tr=0$

We know if a+b+c=0 then a³+b³+c³=3abc

=⟩ {r(s-t)}^3+{s(t-r)}^3+{t(r-s)}^3

$\mathfrak{\blue{= 3{r(s-t)}{s(t-r)}{t(r-s)}}}$

$\mathfrak{\blue{= 3(rs-rt)(st-sr)(tr-ts)}}$

$\huge{\boxed{\bold{\mathbb{\red{HOPE.IT.HELPS}}}}}$