Question

If a4+a2b2+b4 =15 and a2-ab+b2=3, then ab= ?​

Answer 1

Answer:

1

Step-by-step explanation:

Given

a4 + a2b2 + b4 = 8 ----- equation -(i)

And,

a2 + ab + b2 = 4

a2 + b2 = 4 - ab ----------------- equation -(ii)

From Equation (i)

⇒ a4 + b4 + a2b2 = 8

⇒ (a2 + b2)2 - 2a2b2 + a2b2 = 8

⇒ (4-ab)2 - a2b2 = 8

⇒ 16 -8ab + a2b2 - a2b2 = 8

⇒ -8ab = 8 - 16

∴ ab = 1

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Answer 2

Problem type: Factorization

Answer

 We can factorize $\sf{a^4+a^2b^2+b^4}$ with identity.

 If 'zero is added' we have $\sf{a^4+2a^2b^2+b^4-a^2b^2}$.

Given: $\sf{a^4+a^2b^2+b^4}$

$\implies \sf{a^4+2a^2b^2+b^4-a^2b^2}$

$\implies \sf{(a^2+b^2)^2-(ab)^2}$

$\implies \sf{(a^2+ab+b^2)(a^2-ab+b^2)}$

 The second factor is given in the question. If we substitute it in we get another equation.

$\implies\sf{3(a^2+ab+b^2)=15}$

$\implies\sf{a^2+ab+b^2=5}$

 

 Now we have two equations.

• $\sf{a^2+ab+b^2=5}$
• $\sf{a^2-ab+b^2=3}$

 

 Subtract the equations and we have $\sf{2ab=2}$.

$\therefore\sf{ab=1}$

 

More information:

 $\sf{a^4+a^2b^2+b^4}$ is factorized by two identities. Try adding $\sf{a^2b^2-a^2b^2}$ then factorize.

1. $\sf{A^2+2AB+B^2=(A+B)^2}$
2. $\sf{A^2-B^2=(A+B)(A-B)}$