Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
28 Algebraic Identities (Internal Choice)
If \(a+b+c=5\) and \(ab+bc+ca=10\), then prove that \(a^3+b^3+c^3-3abc=-25\).
OR
Factor using the suitable identities:
(A) \(\frac{p^2}{16} - 2 + \frac{16}{p^2}\)
(B) \(9a^2+4b^2+c^2-12ab+6ac-4bc\)
(A) \(\frac{p^2}{16} - 2 + \frac{16}{p^2}\)
(B) \(9a^2+4b^2+c^2-12ab+6ac-4bc\)
Concept:
1
Main: Use the identity \(a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))\). You will first need to find the value of \(a^2+b^2+c^2\) using the expansion of \((a+b+c)^2\).
2
OR: Part A uses the perfect square identity \((x-y)^2 = x^2 - 2xy + y^2\). Part B uses the trinomial square identity \((x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx\).
www.maheshkumarsir.in
Hint:
1
Main: Step 1: Expand \((a+b+c)^2 = 5^2\) to find \(a^2+b^2+c^2\). Step 2: Plug everything into the main cubic identity formula.
2
OR (Part B): Look at the negative signs. The terms \(-12ab\) and \(-4bc\) are negative, while \(+6ac\) is positive. Which variable is common to both negative terms? That variable must carry the negative sign!
www.maheshkumarsir.in
Mistakes to Avoid:
!
Sign Errors in Trinomials: For the OR (Part B), writing \((3a+2b-c)^2\) instead of \((3a-2b+c)^2\). Always check which term causes the specific negative products. Since 'b' is in both negative terms, 'b' must be the negative variable.
www.maheshkumarsir.in
MAIN QUESTION SOLUTION:
Given: \( a+b+c = 5 \) and \( ab+bc+ca = 10 \)
First, we find \( a^2+b^2+c^2 \) using the identity:
\( (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \)
\( (5)^2 = a^2+b^2+c^2 + 2(10) \)
\( 25 = a^2+b^2+c^2 + 20 \)
\( a^2+b^2+c^2 = 25 - 20 = 5 \) [1 M]
Now use the main identity:
\( a^3+b^3+c^3-3abc = (a+b+c)[a^2+b^2+c^2 - (ab+bc+ca)] \) [1 M]
Substitute the known values:
\( a^3+b^3+c^3-3abc = (5)[5 - 10] \)
\( a^3+b^3+c^3-3abc = (5)[-5] \)
\( a^3+b^3+c^3-3abc = -25 \) [1 M]
Hence Proved.
OR
OR QUESTION SOLUTION:
(A) Factorise: \( \frac{p^2}{16} - 2 + \frac{16}{p^2} \)
Rewrite as perfect squares:
\(= \left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2 \)
Using identity \( x^2 - 2xy + y^2 = (x-y)^2 \):
\(= \left(\frac{p}{4} - \frac{4}{p}\right)^2 \) [1.5 M]
(B) Factorise: \( 9a^2+4b^2+c^2-12ab+6ac-4bc \)
Rewrite as perfect squares, noting that terms with 'b' are negative:
\(= (3a)^2 + (-2b)^2 + (c)^2 + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c) \)
Using identity \( (x+y+z)^2 \):
\(= (3a - 2b + c)^2 \) [1.5 M]
29 Graph of Linear Equation
Draw the graph of the linear equation \(x+2y=8\) and find the point on the graph where abscissa is twice the value of ordinate.
Concept:
1
To draw a graph, find at least two or three \((x, y)\) coordinate pairs that satisfy the equation, plot them, and draw a straight line through them.
2
"Abscissa" means the x-coordinate, and "Ordinate" means the y-coordinate. If the abscissa is twice the ordinate, that means \(x = 2y\).
www.maheshkumarsir.in
Hint:
1
Find the intercepts for easy plotting. Let \(y=0\), find \(x\). Let \(x=0\), find \(y\).
2
To find the specific point, substitute \(x = 2y\) directly into the given equation \(x + 2y = 8\) and solve for \(y\).
www.maheshkumarsir.in
Mistakes to Avoid:
!
Swapping Abscissa/Ordinate: Setting \(y = 2x\) instead of \(x = 2y\). This will give you fractional coordinates that are completely incorrect for this specific condition.
www.maheshkumarsir.in
Given equation: \( x + 2y = 8 \)
Let's find coordinates to plot the graph:
If \( y = 0 \), then \( x + 0 = 8 \implies x = 8 \). Point: \((8, 0)\)
If \( y = 4 \), then \( x + 8 = 8 \implies x = 0 \). Point: \((0, 4)\)
If \( y = 2 \), then \( x + 4 = 8 \implies x = 4 \). Point: \((4, 2)\)
[1.5 M for Graph]
Condition: Abscissa is twice the value of ordinate.
This means \( x = 2y \).
Substitute \( x = 2y \) into the equation \( x + 2y = 8 \):
\( 2y + 2y = 8 \)
\( 4y = 8 \implies y = 2 \)
Substitute \( y = 2 \) back to find \( x \):
\( x = 2(2) = 4 \)
The required point on the graph is \( (4, 2) \). [1.5 M]
Point = (4, 2)
Disclaimer: This content is provided solely for educational purposes to help students with their exam preparations. While we strive for maximum accuracy, this guide does not claim absolute authenticity and has no official affiliation with the school's exam setting authorities. It is a preparatory guide created upon student requests. For any discrepancies or issues regarding the content, please contact Maheshkumar Sir, and it will be modified or removed immediately.
