krit.club logo

Exploring Algebraic Identities - Finding New Identities

Grade 9CBSE

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

🔑Concepts

•

An algebraic identity is an algebraic equation that is true for all values of the variables occurring in it.

•

The square of a trinomial is given by (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx. This identity can be extended to any number of terms.

•

Cubic expansions for binomials include (x+y)3(x + y)^3 and (x−y)3(x - y)^3, which are useful for simplifying expressions involving cubes of sums or differences.

•

New identities for factoring cubes, such as x3+y3x^3 + y^3 and x3−y3x^3 - y^3, are derived from the cubic expansion identities.

•

The 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) is a powerful tool for factorizing complex expressions.

•

Conditional Identity: If x+y+z=0x + y + z = 0, then it follows that x3+y3+z3=3xyzx^3 + y^3 + z^3 = 3xyz.

📐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)(x + y)^3 = x^3 + y^3 + 3xy(x + y)

(x−y)3=x3−y3−3xy(x−y)(x - y)^3 = x^3 - y^3 - 3xy(x - y)

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 (3a+4b+5c)2(3a + 4b + 5c)^2.

Solution:

Using the identity (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx, let x=3ax = 3a, y=4by = 4b, and z=5cz = 5c. (3a+4b+5c)2=(3a)2+(4b)2+(5c)2+2(3a)(4b)+2(4b)(5c)+2(5c)(3a)(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a) =9a2+16b2+25c2+24ab+40bc+30ca= 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca

Explanation:

Apply the trinomial square identity by substituting the specific terms for xx, yy, and zz and simplifying the coefficients.

Problem 2:

Factorize 64m3−343n364m^3 - 343n^3.

Solution:

Recognize the expression as a difference of cubes: 64m3=(4m)364m^3 = (4m)^3 and 343n3=(7n)3343n^3 = (7n)^3. Use the identity x3−y3=(x−y)(x2+xy+y2)x^3 - y^3 = (x - y)(x^2 + xy + y^2) with x=4mx = 4m and y=7ny = 7n. 64m3−343n3=(4m−7n)((4m)2+(4m)(7n)+(7n)2)64m^3 - 343n^3 = (4m - 7n)((4m)^2 + (4m)(7n) + (7n)^2) =(4m−7n)(16m2+28mn+49n2)= (4m - 7n)(16m^2 + 28mn + 49n^2)

Explanation:

Identify the terms as perfect cubes and apply the difference of cubes formula.

Problem 3:

Evaluate (−12)3+73+53(-12)^3 + 7^3 + 5^3 without actually calculating the cubes.

Solution:

Let a=−12a = -12, b=7b = 7, and c=5c = 5. First, check the sum: −127+50\begin{array}{r} -12 \\ 7 \\ + 5 \\ \hline 0 \end{array} Since a+b+c=−12+7+5=0a + b + c = -12 + 7 + 5 = 0, we use the conditional identity a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc. (−12)3+73+53=3(−12)(7)(5)(-12)^3 + 7^3 + 5^3 = 3(-12)(7)(5) =3(−420)= 3(-420) =−1260= -1260

Explanation:

When the sum of three numbers is zero, the sum of their cubes equals three times their product.