An algebra problem by Kshetrapal Dashottar

Algebra Level 4

$\begin{cases} a^3 = 3 (b^2+c^2)-25 \\ b^3 = 3(c^2+a^2) -25 \\ c^3 = 3 (a^2 + b^2) - 25 \end{cases}$

Let $a,b$ and $c$ be distinct real numbers satisfying the system of equations above. Find the product $abc$ .

2 3 28 2.5 none of these

1 solution

Kshetrapal Dashottar
Jun 21, 2016

(a) a3 – b3 = 3(b2 – a2)  a2 + b2 + ab = – 3 (a + b) ..(i)
b3 – c3 = 3 (c2 – b2)  b2 + c2 + bc = – 3 (c + b) ..(ii)
a3 – c3 = 3 (c2 – a2)  a2 + c2 + ac = –3 (a + c) ..(iii)
from (i) & (ii)
a2 – c2 + b (a – c) = – 3 (a – c)
a + b + c = – 3 ..(iv)
by adding given equation
a3 + b3 + c3 = 6 (a2 + b2 + c2) – 75
(a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc = 6 (a2 + b2 + c2) – 75
– 3a2 – 3b2 – 3c2 + 3ab + 3bc + 3ac + 3abc = 6(a2 + b2 + c2) – 75
3ab + 3bc + 3ac + 3abc = 9(a2 + b2 + c2) – 75
= 9[(a + b + c)2 – 2ab – 2bc – 2ca)] – 75
= 9[9 – 2ab – 2bc – 2ac) – 75
3ab + 3bc + 3ac + 3abc = 6 – 18 (ab + bc + ca)
3abc = 6 – 21 (ab + bc + ca)
abc = 2 – 7 (ab + bc + ca) ..(v)
from (i), (ii),(iii)
2a2 + 2b2 + 2c2 + ab + bc + ac = – 3  2  – 3 = 18
2(a2 + b2 + c2) + ab + bc + ca = 18
2((a + b + c)2 – 2ab – 2bc – 2ca) + ab + bc + ca = 18
2(3)2 – 3 (ab + bc + ca) = 18
ab + bc + ca = 0 ..(vi)
 from (v) & (vi)
abc = 2 – 7(0) = 2

Did you write NMTC-2015 final test?

Ayush G Rai - 4 years, 10 months ago

i am nmtc stage 2 gold medalist

Kshetrapal Dashottar - 4 years, 9 months ago

An algebra problem by Kshetrapal Dashottar

1 solution

0 pending reports