Question

If p = 4 – 9, prove that p3 + q3 + 12pq = 64

Answer 1

Question:  (with correction)

If p = 4 – q, prove that p³ + q³+ 12pq = 64

Given:

p = 4 - q

To prove:

p³ + q³ + 12pq = 64

Solution :

There are two ways of solving this question.

•  First method  

p = 4 - q  

Now we will substitute the value of p in p³ + q³ +12pq.  

=> (4-q)³ + q³ + 12 (4-q)(q)  

expanding (4-q)³ using algebraic identity  

{ (a-b)³ = a³ - b³ -3a²b +3ab² }  

=> (4)³ - q³ -3(4)²q +3(4)q²  +q³ +12 (4-q)(q)

=> 64  -3(16)q +12q² + 12(4q-q²)

=> 64 - 48q + 12q² + 48q -12q²

=> 64  

Hence proved  

• Second method  

p = 4 - q

p + q = 4  

Cubing both sides  (p+q)³ = (4)³  

We will expand (p+q)³ using algebraic identity  

{ (a+b)³ = a³ + b³ +3ab(a+b) }  

=> (p+q)³ = (4)³

=> p³ + q³ + 3pq(p+q) = 64  

Now we have p+q = 4

=> p³ + q³ +3(4)pq = 64

=> p³ + q³ + 12pq = 64  

Hence proved