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If a + b + c = 5 and ab + bc + ca = 10, then prove that a cube + b cube + c cube – 3abc = -25.
We utilize the algebraic identity for a three-variable cube expression. The identity states that the target expression equals (a + b + c) multiplied by the quantity (a square + b square + c square - ab - bc - ca). We first square the addition equation to get 25 = (a square + b square + c square) + 2Read more
We utilize the algebraic identity for a three-variable cube expression. The identity states that the target expression equals (a + b + c) multiplied by the quantity (a square + b square + c square – ab – bc – ca). We first square the addition equation to get 25 = (a square + b square + c square) + 2(10), meaning the squared sum equals 5. Plugging these values back into our main identity yields 5 times (5 – 10), which equals minus 25.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessIf both x – 2 and x – 1/2 are factors of px square + 5x + r, show that p = r.
Since the given binomial equations are exact factors, we apply the factor theorem to solve the unknown coefficients. Setting x equal to 2 yields the equation 4p + 10 + r = 0. Setting x equal to 1/2 yields p/4 + 5/2 + r = 0, which multiplies out to p + 10 + 4r = 0. Equating both zero expressions giveRead more
Since the given binomial equations are exact factors, we apply the factor theorem to solve the unknown coefficients. Setting x equal to 2 yields the equation 4p + 10 + r = 0. Setting x equal to 1/2 yields p/4 + 5/2 + r = 0, which multiplies out to p + 10 + 4r = 0. Equating both zero expressions gives 4p + 10 + r = p + 10 + 4r. Cancelling 10 and rearranging the terms simplifies to 3p = 3r, proving p = r.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessA rectangular pool has area 2x square + 7x + 3 square hastas
The area of any regular rectangle is found by multiplying its length by its width dimensions. Therefore, to calculate the missing length expression, we must divide the total area polynomial by the width expression. Factorizing the area polynomial 2x square + 7x + 3 by middle term splitting gives theRead more
The area of any regular rectangle is found by multiplying its length by its width dimensions. Therefore, to calculate the missing length expression, we must divide the total area polynomial by the width expression. Factorizing the area polynomial 2x square + 7x + 3 by middle term splitting gives the paired factors (2x + 1) multiplied by (x + 3). Cancelling out the common width binomial factor (2x + 1) leaves the remaining linear factor (x + 3).
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessIf a number plus its reciprocal equals 10/3, find the number.
Let the unknown target number be represented by the variable x. According to the wording of the problem, the algebraic equation is written as x plus 1/x equals 10/3. Multiplying the entire equation by 3x clears out the denominators, transforming it into the standard quadratic equation form 3x squareRead more
Let the unknown target number be represented by the variable x. According to the wording of the problem, the algebraic equation is written as x plus 1/x equals 10/3. Multiplying the entire equation by 3x clears out the denominators, transforming it into the standard quadratic equation form 3x square – 10x + 3 = 0. Splitting the middle term results in the linear factors (3x – 1) and (x – 3), giving the solutions 3 or 1/3.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessThe village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
The inner playground is a square with an area of 40 times 40, which equals 1600 square metres. When a path of width s is added around all sides, the new outer boundary forms a larger square with a total side length of 40 + 2s metres. The total outer area is the square of (40 + 2s), which expands toRead more
The inner playground is a square with an area of 40 times 40, which equals 1600 square metres. When a path of width s is added around all sides, the new outer boundary forms a larger square with a total side length of 40 + 2s metres. The total outer area is the square of (40 + 2s), which expands to 1600 + 160s + 4s square. Subtracting the inner playground area leaves the path area.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
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