Q. 10 D4.4( 5 Votes )
Find the product:
Answer :
First solving 
Using the identity (a +b) (a–b) = a2 – b2
Here 
Now solving 
Again, Using the identity (a +b) (a–b) = a2 – b2
Here 
= 
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Find the product:
(2x – 3y)(2x + 3y)(4x2 + 9y2)
Karnataka Board - Mathematics Part IIf a, b are rational numbers such that a2 + b2 + c2 – ab – bc – ca = 0, prove that a = b = c.
Karnataka Board - Mathematics Part IUse the identity (a + b)(a – b) = a2 – b2 to find the products:
(i) (x – 6) (x + 6)
(ii) (3x + 5)(3x + 5)
(iii) (2a + 4b)(2a – 4b)
(iv) 
Express the following as difference of two squares:
(i) (x + 2z)(2x + z);
(ii) 4(x + 2y)(2x + y);
(iii) (x + 98)(x + 102);
(iv) 505 × 495.
Karnataka Board - Mathematics Part IFind the product:
(2a + 3)(2a – 3)(4a2 + 9)
Karnataka Board - Mathematics Part ISimplify:
(i) (x + y)2 + (x – y)2;
(ii) (x + y)2 × (x – y)2.
Karnataka Board - Mathematics Part IFind the product:
(p + 2) (p – 2)(p2 + 4)
Karnataka Board - Mathematics Part IFind the product:
(x – 3)(x + 3)(x2 + 9)
Karnataka Board - Mathematics Part IEvaluate these using identity:
(i) 55 × 45 (ii) 33 × 27
(iii) 8.5 × 9.5 (iv) 102 × 98
Karnataka Board - Mathematics Part IFind the product:
