Question

(3a-2b)*(9a2+4b2+6ab)
the 2 in 9a2 and 4b2 is square

Answer 1

Answer:

Step 1: Draw a square ACDF with AC=3a.

Step 2: Cut AB=2b, so that BC=(3a−2b).

Step 3: Complete the squares and rectangle as shown in the diagram.

Step 4: Area of yellow square IDEO= Area of square ACDF− Area of rectangle GOFE− Area of rectangle BCIO− Area of red square ABOG

Therefore, (3a−2b)  

2

=(3a)  

2

−2b(3a−2b)−2b(3a−2b)−(2b)  

2

 

= 9a  

2

−6ab+4b  

2

−6ab+4b  

2

−4b  

2

 

= 36x  

2

+5  

2

−60x

Hence, geometrically we proved the identity (3a−2b)  

2

=9a  

2

+4b  

2

−12ab.

Step-by-step explanation: