SOLVING PROBLEMS USING ALGEBRAIC IDENTITIES

Problem 1 :

Expand : (2x + 3y)²

Solution :

We have 

(a + b)² = a² + 2ab + b²

Substitute a = 2x and b = 3y.

(2x + 3y)² = (2x)² + 2(2x)(3y) + (3y)²

= 2²x² + 2(2x)(3y) + 3²y²

= 4x² + 12xy + 9y²

Problem 2 :

Expand : (4a - 5b)²

Solution :

We have

(a - b)² = a² - 2ab + b²

Substitute a = 4a and b = 5b.

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

= 4²a² - 40ab + 5²b²

= 16p² - 40ab + 25b²

Problem 3 :

Expand : (5p + 4q)(5p - 4q)

Solution :

We have 

(a + b)(a - b) = a² - b²

Substitute a = 5p and b = 4q.

(5p + 4q)(5p - 4q) = (5p)² - (4q)²

= 5²p² - 4²q²

= 25p² - 16q²

Problem 4 :

Expand (3x - 4y)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 3x, and b = -4y.

(3x - 4y)³ = (3x)³ + 3(3x)²(-4y) + 3(3p)(-4q)² + (-4y)³

= 27x³ + 3(9x²)(-4y) + 3(3x)(16y²) + (-64y³)

= 27x³ - 108x²y + 144xy² - 64y³

Problem 5 :

Expand (x + 1/y)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = x, and b = 1/y.

(x + 1/y)³ = x³ + 3(x)²(1/y) + 3(x)(1/y)² + (1/y)³

= x³ + 3x²/y + 3x/y² + 1/y³

Problem 6 :

Evaluate using identity : 98³.

Solution :

98³ = (100 - 2)³

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 100, and b = -2.

(100 - 2)³ = 100³ + 3(100)²(-2) + 3(100)(-2)² + (-2)³

98³ = 1000000 + 3(10000)(-2) + 3(100)(4) - 8

= 1000000 - 60000 + 1200 - 8

= 941192

Problem 7 :

Evaluate using identity : 1001³.

Solution :

1001³ = (1000 + 1)³

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 1000, and b = 1.

(1000 + 1)³ = 1000³ + 3(1000)²(1) + 3(1000)(1)² + (1)³

(1001)³ = 1000³ + 3(1000)²(1) + 3(1000)(1)² + (1)³

= 1000000000 + 3(1000000)(1) + 3(1000)(1) + 1

= 1000000000 + 3000000 + 3000 + 1

= 1003003001

Problem 8 :

Find 8x³ + 27y³, if 2x + 3y = 13 and xy = 6.

Solution :

We know that

a³ + b³ = (a + b)³ - 3ab(a + b)

Substitute a = 2x, and b = 3y.

(2x)³ + (3y)³ = (2x + 3y)³ - 3(2x)(3y)(2x + 3y)

8x³ + 27y³ = (2x + 3y)³ - 18(xy)(3x + 4y)

Substitute 2x + 3y = 13 and xy = 6.

8x³ + 27y³ = (13)³ - 18(6)(13)

= 2197 - 1404

= 793

Problem 9 :

Expand : (p + 2q + 3)²

Solution :

We know that

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

Substitute a = p, b = 2q and c = 3.

(p + 2q + 3)² = p² + (2q)² + 3² + 2[(p)(2q) + (2q)(3) + (p)(3)]

= p² + 4q² + 9 + 2[2pq + 6q + 3p]

= p² + 4q² + 9 + 4pq + 12q + 6p

= p² + 4q² + 6p + 12q + 4pq + 9

Problem 10 :

Expand : (x + 2y - 3z)²

Solution :

We know that

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

Substitute a = x, b = 2y and c = -3z.

(x + 2y - 3z)²

= x² + (2y)² + (-3z)² + 2[(x)(2y) + (2y)(-3z) + (x)(-3z)]

= x² + 4y² + 9z² + 2[2xy - 6yz - 3xz]

= x² + 4y² + 9z² + 4xy - 12yz - 6xz

Problem 11 :

If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x² + y² + z².

Solution :

We know that

(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)

Then,

x² + y² + z² = (x + y + z)² - 2(xy + yz + xz)

Substitute (x + y + z) = 9 and (xy + yz + zx) = 26.

x² + y² + z² = 9² - 2(26)

=  81 - 52

= 29

Problem 12 :

If (x + a)(x + b)(x + c) = x³ - 10x² + 45x - 15, then find the values of the following.

(i) a + b + c

(ii) 1/a +1/b + 1/c

(iii) a² + b² + c²

Solution :

(x + a)(x + b)(x + c) = x³ - 10x² + 45x - 15

x³ + (a + b + c)x² + (ab + bc + ca)x + abc = x³- 10x² + 45x - 15

Comparing the coefficients of x², x and constant term,

a + b + c = -10

ab + bc + ca  =  45

abc = -15

(i) a + b + c :

a + b + c = -10

(ii) 1/a + 1/b + 1/c :

Least common multiple of (a, b, c) = abc.

1/a + 1/b + 1/c = (bc/abc) + (ac/abc) + (ab/abc)

= (ab + bc ac)/abc

Substitute (ab + bc + ac) = 45 and abc = -15.

= 45/(-15)

= -3

(ii) a² + b² + c² :

a² + b² + c² = (a + b + c)² - 2(ab + bc + ac)

Substitute (a + b + c) = -10 and (ab + bc + ac) = 45.

= (-10)² - 2(45)

= 100 - 90

= 10

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