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Exploring Algebraic Identities - Discover and apply additional algebraic identities across varied expressions

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Expansion of a Trinomial Square: The identity (x+y+z)2(x + y + z)^2 expands to the sum of the squares of each term plus twice the product of each possible pair of terms.

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Cubes of Binomials: Expanding expressions like (x+y)3(x + y)^3 and (x−y)3(x - y)^3 results in four-term polynomials using the coefficients from Pascal's triangle (1, 3, 3, 1).

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Sum and Difference of Cubes: Factoring x3+y3x^3 + y^3 and x3−y3x^3 - y^3 into a linear factor and a quadratic factor.

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The Three-Variable Cubic Identity: Applying the complex identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx).

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Conditional Identity: A special case where if x+y+z=0x + y + z = 0, then the expression x3+y3+z3x^3 + y^3 + z^3 simplifies directly to 3xyz3xyz.

📐Formulae

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

(x+y)3=x3+y3+3xy(x+y)=x3+3x2y+3xy2+y3(x + y)^3 = x^3 + y^3 + 3xy(x + y) = x^3 + 3x^2y + 3xy^2 + y^3

(x−y)3=x3−y3−3xy(x−y)=x3−3x2y+3xy2−y3(x - y)^3 = x^3 - y^3 - 3xy(x - y) = x^3 - 3x^2y + 3xy^2 - y^3

x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2)

x3−y3=(x−y)(x2+xy+y2)x^3 - y^3 = (x - y)(x^2 + xy + y^2)

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

If x+y+z=0, then x3+y3+z3=3xyz\text{If } x + y + z = 0, \text{ then } x^3 + y^3 + z^3 = 3xyz

💡Examples

Problem 1:

Expand (2x−3y+4z)2(2x - 3y + 4z)^2 using algebraic identities.

Solution:

(2x−3y+4z)2=(2x)2+(−3y)2+(4z)2+2(2x)(−3y)+2(−3y)(4z)+2(4z)(2x)=4x2+9y2+16z2−12xy−24yz+16zx(2x - 3y + 4z)^2 = (2x)^2 + (-3y)^2 + (4z)^2 + 2(2x)(-3y) + 2(-3y)(4z) + 2(4z)(2x) = 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx

Explanation:

Applied the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca, substituting a=2xa = 2x, b=−3yb = -3y, and c=4zc = 4z.

Problem 2:

Evaluate 99399^3 using a suitable identity.

Solution:

993=(100−1)3=1003−13−3(100)(1)(100−1)=1000000−1−300(99)=1000000−1−29700=97029999^3 = (100 - 1)^3 = 100^3 - 1^3 - 3(100)(1)(100 - 1) = 1000000 - 1 - 300(99) = 1000000 - 1 - 29700 = 970299. Calculation check: 1000000−29701970299\begin{array}{r} 1000000 \\ - 29701 \\ \hline 970299 \end{array}

Explanation:

Used the identity (x−y)3=x3−y3−3xy(x−y)(x - y)^3 = x^3 - y^3 - 3xy(x - y) by splitting 9999 as (100−1)(100 - 1) to simplify the calculation.

Problem 3:

Factorize 8x3+y3+27z3−18xyz8x^3 + y^3 + 27z^3 - 18xyz.

Solution:

8x3+y3+27z3−18xyz=(2x)3+(y)3+(3z)3−3(2x)(y)(3z)8x^3 + y^3 + 27z^3 - 18xyz = (2x)^3 + (y)^3 + (3z)^3 - 3(2x)(y)(3z). Using the identity, we get (2x+y+3z)((2x)2+y2+(3z)2−(2x)(y)−(y)(3z)−(3z)(2x))=(2x+y+3z)(4x2+y2+9z2−2xy−3yz−6zx)(2x + y + 3z)((2x)^2 + y^2 + (3z)^2 - (2x)(y) - (y)(3z) - (3z)(2x)) = (2x + y + 3z)(4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6zx).

Explanation:

Identified the expression as the form a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc where a=2xa = 2x, b=yb = y, and c=3zc = 3z.

Problem 4:

Without actual cubing, find the value of (−12)3+73+53(-12)^3 + 7^3 + 5^3.

Solution:

Let a=−12,b=7,c=5a = -12, b = 7, c = 5. Sum a+b+c=−12+7+5=0a + b + c = -12 + 7 + 5 = 0. Since a+b+c=0a + b + c = 0, a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc. Value =3(−12)(7)(5)=−1260= 3(-12)(7)(5) = -1260.

Explanation:

Used the conditional identity: if x+y+z=0x + y + z = 0, then x3+y3+z3=3xyzx^3 + y^3 + z^3 = 3xyz.