Question

if a,b,c are a continued proportion, then prove that abc (a+b+c) whole cube= (ab+bc+ca)ka whole cube

Answer 1

abc(a + b + c)³ =  (ab + bc + ca)³ if a,b,c are a continued proportion

Step-by-step explanation:

a,b,c are a continued proportion

=> b = ak

& c = bk  = (ak)k = ak²

LHS

= abc(a + b + c)³

= a(ak)(ak²)(a + ak + ak²)³

= a³k³ a³(1 + k + k²)³

= a⁶k³(1 + k + k²)³

RHS

= (ab + bc + ca)³

= (aak + akak² + ak²a)³

= (a²k + a²k³ + a²k)³

= (a²k(1 + k² + k))³

= (a²k)³(1 + k + k²)³

=  a⁶k³(1 + k + k²)³

LHS = RHS

QED

Proved

abc(a + b + c)³ =  (ab + bc + ca)³

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

a,b,c are a continued proportion

=> b = ak

& c = bk = (ak)k = ak²

LHS

= abc(a + b + c)³

= a(ak)(ak²)(a + ak + ak²)³

= a³k³ a³(1 + k + k²)³

= a⁶k³(1 + k + k²)³

RHS

= (ab + bc + ca)³

= (aak + akak² + ak²a)³

= (a²k + a²k³ + a²k)³

= (a²k(1 + k² + k))³

= (a²k)³(1 + k + k²)³

= a⁶k³(1 + k + k²)³

LHS = RHS

QED

Proved

abc(a + b + c)³ = (ab + bc + ca)³