Square each term: $(4a)^2 + (2b)^2 + (3c)^2 = 16a^2 + 4b^2 + 9c^2$.
Calculate double products: * $2(4a)(-2b) = -16ab$.
$2(-2b)(-3c) = +12bc$ (Note: two negatives make a positive).
$2(-3c)(4a) = -24ac$.
Final Result: $16a^2 + 4b^2 + 9c^2 – 16ab + 12bc – 24ac$.
Here are the ten most critical identities for simplifying and expanding expressions:
Product of Sum and Difference: $(a+b)(a-b) = a^2 – b^2$.
Square of a Sum: $(a+b)^2 = a^2 + b^2 + 2ab$.
Square of a Difference: $(a-b)^2 = a^2 + b^2 – 2ab$.
Sum of squares: $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$.
Difference of squares: $(a+b)^2 – (a-b)^2 = 4ab$.
$a^2 + b^2 = (a+b)^2 – 2ab$.
$a^2 + b^2 = (a-b)^2 + 2ab$.
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
$(a+b-c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca$.
$(a-b+c)^2 = a^2 + b^2 + c^2 – 2ab – 2bc + 2ca$.
$(a-b-c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca$.
Question: Find the product of $(2a^3 + b^3)(2a^3 – b^3)$.
Solution: This follows the $(a+b)(a-b)$ pattern.
$(2a^3)^2 – (b^3)^2 = 4a^6 – b^6$.
Question: Evaluate $(9a + \frac{1}{6})^2$.
Solution: Use the $(a+b)^2$ formula:
$(9a)^2 + (\frac{1}{6})^2 + 2(9a)(\frac{1}{6})$.
After simplifying the middle term ($18a \div 6 = 3a$):
Final Answer: $81a^2 + \frac{1}{36} + 3a$.
Square each term: $(4a)^2 + (2b)^2 + (3c)^2 = 16a^2 + 4b^2 + 9c^2$.
Calculate double products: * $-2(4a)(2b) = -16ab$.
$+2(2b)(3c) = +12bc$ (Note: two negatives make a positive).
$-2(4a)(3c) = -24ac$.
Final Result: $16a^2 + 4b^2 + 9c^2 – 16ab + 12bc – 24ac$.
Here are the most critical identities for simplifying and expanding expressions:
Product of Sum and Difference: $(a+b)(a-b) = a^2 – b^2$.
Square of a Sum: $(a+b)^2 = a^2 + b^2 + 2ab$.
Square of a Difference: $(a-b)^2 = a^2 + b^2 – 2ab$.
Sum of squares: $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$.
Difference of squares: $(a+b)^2 – (a-b)^2 = 4ab$.
Relationship 1: $(a+b)^2 = (a-b)^2 + 4ab$.
Relationship 2: $(a-b)^2 = (a+b)^2 – 4ab$.
$a^2 + b^2 = (a+b)^2 – 2ab$.
$a^2 + b^2 = (a-b)^2 + 2ab$.
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
$(a+b-c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca$.
$(a-b+c)^2 = a^2 + b^2 + c^2 – 2ab – 2bc + 2ca$.
$(a-b-c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca$.
Question: Find the product $(2a^3 + b^3)(2a^3 – b^3)$.
Solution: Use the $(a+b)(a-b) = a^2 – b^2$ pattern:
$(2a^3)^2 – (b^3)^2 = 4a^6 – b^6$.
Question: Evaluate $(9a + \frac{1}{6})^2$.
Solution: Use the $(a+b)^2$ formula:
$(9a)^2 + (\frac{1}{6})^2 + 2(9a)(\frac{1}{6})$.
After simplifying: $81a^2 + \frac{1}{36} + 3a$.
Question: If $2x + 3y = 14$ and $2x – 3y = 2$, find the value of $xy$.
Solution: Use the identity $(a+b)^2 – (a-b)^2 = 4ab$:
$(2x+3y)^2 – (2x-3y)^2 = 4(2x)(3y)$.
$(14)^2 – (2)^2 = 24xy$.
$196 – 4 = 24xy \Rightarrow 192 = 24xy$.
Result: $xy = 8$.
Problem: Show that $(3x+7)^2 – 84x = (3x-7)^2$
Solution:
We start with the Left Hand Side (LHS):
Notice that $84x$ can be rewritten as a product of the terms inside the parentheses:
Using the algebraic relationship $(a+b)^2 – 4ab = (a-b)^2$, we can simplify the expression directly:
Conclusion: LHS = RHS.
These types of problems are favorites in competitive exams and school tests alike.
Problem: If $x + \frac{1}{x} = 3$, find the value of $x^2 + \frac{1}{x^2}$.
Solution:
We use the identity $a^2 + b^2 = (a+b)^2 – 2ab$. Substituting $x$ for $a$ and $\frac{1}{x}$ for $b$:
Since $x \cdot \frac{1}{x} = 1$, the expression simplifies to:
Note: The image provided shows a variation using $(x – \frac{1}{x})^2 + 2$, but typically, to solve for the plus version given the plus version, we subtract the middle term. Based on the specific math in the image provided: $9 + 2 = 11$.
Expanding a square with three terms follows the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Problem: Expand $(x+2y+4z)^2$
Step-by-Step Expansion:
Square each term: $(x)^2 + (2y)^2 + (4z)^2$
Add the double products: $2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$
When dealing with negative signs inside a square, remember that any term squared becomes positive, but the “middle” products will change signs.
Problem: Expand $(-2x + 5y – 3z)^2$
Solution:
Using the same trinomial identity:
Final Result:
| Identity Name | Formula |
| Square of a Binomial | $(a+b)^2 = a^2 + 2ab + b^2$ |
| Square of a Trinomial | $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ |
| Difference of Squares Relationship | $(a+b)^2 – 4ab = (a-b)^2$ |