An Elegant System of Equations

at first , $a^2 = bc$ is given >> then $abc = a^3$

$(a+b+c)^2=(56)^2$

$a^2 + b^2 +c^2 + 2ab+2bc+2ca=3136$

$1344+2(b+c)+2a^2=3136$

$1344+2a(a+b+c)=3136$

$1344+2a*56=3136$

$a=16$

then $abc = a^3=(16)^3=4096$

$a+b+c=56\quad \quad \quad \quad \quad \rightarrow \quad 1\\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=1344\quad \rightarrow \quad 2\\ { a }^{ 2 }=bc\quad \quad \quad \quad \quad \quad \quad \quad \quad \rightarrow \quad 3\\ from\quad equation1,\quad we\quad know\quad that\quad b+c=56-a\quad \quad \rightarrow 4\\ and\quad equation2\quad can\quad be\quad written\quad as\quad { (b+c) }^{ 2 }-{ a }^{ 2 }=1344\quad \quad \rightarrow 5\quad ,since\quad { a }^{ 2 }=bc\\ substitute\quad equation4\quad into\quad equation5\quad we\quad get\quad { (56-a) }^{ 2 }-{ a }^{ 2 }=1344\\ after\quad some\quad algebraic\quad manipulations\quad we\quad get\quad a=16\\ substitute\quad it\quad in\quad equation3\quad we\quad get\quad bc=256\\ thus,\quad abc=4096\\ \quad \quad \quad \quad \quad \quad \quad \quad \\$

$(a+b+c)^{2}-(a^{2}+b^{2}+c^{2}) = 2(ab+bc+ac) = 2a(a+b+c)$ where we used that $bc=a^{2}$ . Now using the given values for the expressions in parethesis we have

$56^{2} - 1344 = 2\cdot 56 \cdot a \rightarrow 1792 = 112a \rightarrow a=16.$

But now $abc=a\cdot a^2 = a^3 = 16^3 = 4096$

(a+b+c)^2=(56)^2 => (a)^2+(b)^2+(c)^2+2ac+2bc+2ab=(56)^2 => 1344+2ac+2(a)^2+2ab=(56)^2 => 1344+2a(a+b+c)=(56)^2 => 1344+2a(56)=(56)^2 => a=16 => abc= a (a)^2 => 16 (16)^2 => 4096.

a + b + c = 56

a = 56 - b - c

squaring both sides:

a^2 = (56 – b – c)^2

a^2 = 56^2 + b^2 + c^2 + 2(bc – 56b – 56c)

a^2 = 3136 + b^2 + c^2 + 2bc – 112b – 112c

now

a^2 + b^2 + c^2 = 1344 b^2 + c^2 = 1344 – a^2

so

a^2 = 3136 + 1344 – a^2 + 2bc – 112b – 112c

a^2 = 4480 – a^2 + 2a^2 – 112(b+c)

a^2 + a^2 – 2a^2 = 4480 – 112(b+c)

112(b+c) = 4480

(b+c) = 4480/112

(b+c) = 40

so

a + b + c = 56 a = 56 – 40 a = 16

now

bc = a^2

so

abc = a * bc = a * a^2 = a^3 = 16^3 = 4096

b^2+c^2+2bc-bc=1344 (bc=a^2)

(b+c)^2-a^2=1344 (complete square)

but, b+c=56-a

then, a^2-112a+3136-a^2=1344

simplifing we get, 112a=1792

a=16

bc=16^2

then,abc=16^3

finally, abc=4096.

start with a^2=bc. substitution for a^2 in a^2+b^2+c^2=1344. So b^2+c^2+bc=1344 add bc to both sides. so (b+c)^2 = 1344+bc so then (b+c)^2=1344+bc. Thats the same as (b+c)^2=1344+a^2. a+b+c=56 so b+c=56-a. Then (56-a)^2=1344+a^2. expand and simplify so that 56^2-1344=112a. that simplifies to 1792=112a. a=16. abc=a^3 so abc is a^3 = 16^3 = 4096

a^2 = bc implies that b,a,c are in GP. so b + a + c = b (1 + r + r^2) = 56, where r is the common ratio. using some trial and error and substituting the value in a^2 + b^2 + c^2 = 1344 to verify, it is easy to get r = 2. so b = 8. hence abc = (br)^3 = 16^3 = 4096

a2 + b2 + c2 = 1344 a2 = bc so, bc + b2 + c2 = 1344 by adding ‘bc’ both side, we get 2bc + b2 + c2 =1344 + bc ((b+c)2 = 2bc + b2 + c2 ) (b+c)2 = 1344 + bc (56-a)2 = 1344 + bc (a+b+c=56… so b+c = 56-a) a2 + 3136 -112a =1344 + a2 (a2 = bc given) 112a=3136-1344 a=1792/112 a=16 so, a2=bc a3=abc so, abc = 4096

(a+b+c)^2=3136 (a^2+b^2+c^2)(2a(a+b+c))=3136 _ 1 (a^2+b^2+c^2)=1344 a+b+c= 56 recoupment _ 1 a=bc a=16 , a+b=40 , bc=256

Solution 1.

Label the equations in the prescribed order as (1), (2) and (3). By (1) and (3),

(b+c)^2 = (56-a)^2,

b^2+2bc+c^2 = 3136-112a+a^2,

(4) b^2+bc+c^2 = 3136-112a.

By (2) and (3),

(5) b^2+bc+c^2 = 1344.

Equating (4) and (5),

3136-112a = 1344,

a = 16.

Therefore abc = a(a^2) = a^3 = 4096.

Solution 2.

By the first two equations,

(a+b+c)^2-(a^2+b^2+c^2) = 56^2-1344 = 1792,

ab+bc+ca = 896,

a(b+c)+bc = 896.

Substituting b+c = 56-a and a^2 = bc from the first and third equations,

a(56-a)+a^2 = 896,

a = 16.

Therefore abc = a(a^2) = a^3 = 4096.

a + b + c = 56

(a + b + c)² = 56²

a² + b² + c² + 2(ab + bc+ ac) = 56²

ab + bc + ac = (56² - 1344)/2 = 896

Since bc = a²

ab + a² + ac = 896

a (a + b + c) = 896

a = 896/56 = 16

abc = a * bc = a * a² = a³ = 16³ = 4096