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Find the value of x cube – 8y cube – 36xy – 216, when x = 2y + 6
We analyze the given conditional equation by moving terms to one side, which gives the simplified statement x - 2y - 6 = 0. Now, we rewrite the main polynomial expression using these exact bases, formatting it as x cube + (-2y) cube + (-6) cube minus 3 times x times (-2y) times (-6). According to alRead more
We analyze the given conditional equation by moving terms to one side, which gives the simplified statement x – 2y – 6 = 0. Now, we rewrite the main polynomial expression using these exact bases, formatting it as x cube + (-2y) cube + (-6) cube minus 3 times x times (-2y) times (-6). According to algebraic conditions, when the sum of three base terms is equal to zero, the sum of their individual cubes minus three times their product automatically equals zero. Thus, the answer is 0.
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 lessFind the value of x cube + y cube – 12xy + 64 when x + y = -4
We rearrange the given polynomial expression into the standard conditional cubic identity template. We can rewrite the mathematical statement as x cube + y cube + (4) cube minus 3 times x times y times 4. This matches the identity where the three components are x, y and 4. The identity states that iRead more
We rearrange the given polynomial expression into the standard conditional cubic identity template. We can rewrite the mathematical statement as x cube + y cube + (4) cube minus 3 times x times y times 4. This matches the identity where the three components are x, y and 4. The identity states that if the sum of these three numbers equals zero, the total expression equals zero. Since x + y = -4, the sum x + y + 4 equals zero, making the value 0.
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 lessBy factoring the expression, check that n cube – n is always divisible by 6 for all natural numbers n.
We begin by completely factoring the given cubic expression. Taking out the common term n gives n(n square - 1), which further opens up into the three-part continuous variable expression (n - 1)(n)(n + 1). This product represents three consecutive natural numbers. In any sequence of three consecutivRead more
We begin by completely factoring the given cubic expression. Taking out the common term n gives n(n square – 1), which further opens up into the three-part continuous variable expression (n – 1)(n)(n + 1). This product represents three consecutive natural numbers. In any sequence of three consecutive numbers, at least one number must be even (divisible by 2) and exactly one must be a multiple of 3. Their product is therefore always divisible by 2 times 3, which equals 6.
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 + 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/
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