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Mathematics

For x = 5 and y = 2, verify the following:

(i) (x + y)2 = x2 + 2xy + y2

(ii) (x - y)2 = x2 - 2xy + y2

(iii) x2 - y2 = (x + y)(x - y)

(iv) (x + y)2 = (x - y)2 + 4xy

(v) (x + y)3 = x3 + y3 + 3x2y + 3xy2

Algebra Basics

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Answer

(i) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x + y)2

= (5 + 2)2

= (7)2

= 49

RHS = x2 + 2xy + y2

= (5)2 + 2(5)(2) + (2)2

= 25 + 20 + 4

= 49

Since, LHS = RHS.

Hence, (x + y)2 = x2 + 2xy + y2 is verified for x = 5 and y = 2.

(ii) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x - y)2

= (5 - 2)2

= (3)2

= 9

RHS = x2 - 2xy + y2

= (5)2 - 2(5)(2) + (2)2

= 25 - 20 + 4

= 9

Since, LHS = RHS.

Hence, (x - y)2 = x2 - 2xy + y2 is verified for x = 5 and y = 2.

(iii) Substituting x = 5 and y = 2 in the given result, we get:

LHS = x2 - y2

= (5)2 - (2)2

= 25 - 4

= 21

RHS = (x + y)(x - y)

= (5 + 2)(5 - 2)

= 7 × 3

= 21

Since, LHS = RHS.

Hence, x2 - y2 = (x + y)(x - y) is verified for x = 5 and y = 2.

(iv) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x + y)2

= (5 + 2)2

= (7)2

= 49

RHS = (x - y)2 + 4xy

= (5 - 2)2 + 4(5)(2)

= (3)2 + 40

= 9 + 40

= 49

Since, LHS = RHS.

Hence, (x + y)2 = (x - y)2 + 4xy is verified for x = 5 and y = 2.

(v) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x + y)3

= (5 + 2)3

= (7)3

= 343

RHS = x3 + y3 + 3x2y + 3xy2

= (5)3 + (2)3 + 3(5)2(2) + 3(5)(2)2

= 125 + 8 + 3(25)(2) + 3(5)(4)

= 125 + 8 + 150 + 60

= 343

Since, LHS = RHS.

Hence, (x + y)3 = x3 + y3 + 3x2y + 3xy2 is verified for x = 5 and y = 2.

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