May 2026

Find the Value of k Question: If \[ (a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\} = k(a^3+b^3+c^3-3abc) \] find: \[ k \] Solution: Using identity: \[ (a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\} = 2(a^3+b^3+c^3-3abc) \] Comparing with \[ (a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\} = k(a^3+b^3+c^3-3abc) \] We get: \[ k=2 \] Next Question / Full Exercise

Find x³ + y³ + z³ − 3xyz Question: If \[ x+y+z=5 \] and \[ xy+yz+zx=7 \] find: \[ x^3+y^3+z^3-3xyz \] Solution: Using identity: \[ x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \] First find \[ x^2+y^2+z^2 \] Using identity: \[ (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \] Substituting the given values: \[ 5^2 = x^2+y^2+z^2+2(7) \] \[ 25 = x^2+y^2+z^2+14 \]

Find (x + y)³ + (y + z)³ + (z + x)³ Question: If \[ x+y+z=0 \] find: \[ (x+y)^3+(y+z)^3+(z+x)^3 \] Solution: Since \[ x+y+z=0 \] \[ x+y=-z \] \[ y+z=-x \] \[ z+x=-y \] Therefore, \[ (x+y)^3+(y+z)^3+(z+x)^3 \] \[ =(-z)^3+(-x)^3+(-y)^3 \] \[ =-(x^3+y^3+z^3) \] Now using identity: \[ x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \] Since \[

Find a + b + c Question: If \[ a^2+b^2+c^2=24 \] and \[ ab+bc+ca=-4 \] find: \[ a+b+c \] Solution: Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Substituting the given values: \[ (a+b+c)^2 = 24+2(-4) \] \[ = 24-8 \] \[ =16 \] \[ a+b+c = \sqrt{16} \] \[ =\pm4 \] Next Question / Full

Find a² + 1/a² Question: If \[ a^2-2a-1=0 \] find: \[ a^2+\frac{1}{a^2} \] Solution: Given: \[ a^2-2a-1=0 \] \[ a^2-2a=1 \] Dividing both sides by \[ a \] \[ a-2-\frac{1}{a}=0 \] \[ a-\frac{1}{a}=2 \] Using identity: \[ \left(a-\frac{1}{a}\right)^2 = a^2+\frac{1}{a^2}-2 \] Substituting: \[ (2)^2 = a^2+\frac{1}{a^2}-2 \] \[ 4 = a^2+\frac{1}{a^2}-2 \] \[ a^2+\frac{1}{a^2} =

Evaluate (37³ − 28³)/(37² + 37×28 + 28²) Evaluate: \[ \frac{37^3-28^3}{37^2+37\times28+28^2} \] Solution: Using identity: \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] Here, \[ a=37,\qquad b=28 \] \[ 37^3-28^3 = (37-28)(37^2+37\times28+28^2) \] Substituting in the given expression: \[ \frac{(37-28)(37^2+37\times28+28^2)} {37^2+37\times28+28^2} \] \[ =37-28 \] \[ =9 \] Next Question / Full Exercise

Find a² + 1/a² Question: If \[ a+\frac{1}{a}=-2 \] find: \[ a^2+\frac{1}{a^2} \] Solution: Using identity: \[ \left(a+\frac{1}{a}\right)^2 = a^2+\frac{1}{a^2}+2 \] Substituting the given value: \[ (-2)^2 = a^2+\frac{1}{a^2}+2 \] \[ 4 = a^2+\frac{1}{a^2}+2 \] \[ a^2+\frac{1}{a^2} = 4-2 \] \[ =2 \] Next Question / Full Exercise

If (a-b)² + (b-c)² + (c-a)² = 0 Question: If \[ (a-b)^2+(b-c)^2+(c-a)^2=0 \] then ___________ Solution: We know that the square of any real number is always non-negative. Therefore, \[ (a-b)^2 \ge 0 \] \[ (b-c)^2 \ge 0 \] \[ (c-a)^2 \ge 0 \] Their sum is given as: \[ (a-b)^2+(b-c)^2+(c-a)^2=0 \] This is possible

Find a²/bc + b²/ca + c²/ab Question: If \[ a+b+c=0 \] find: \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} \] Solution: Taking LCM: \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} = \frac{a^3+b^3+c^3}{abc} \] Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Since \[ a+b+c=0 \] \[ a^3+b^3+c^3-3abc=0 \] \[ a^3+b^3+c^3=3abc \] Substituting: \[ \frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} \] \[ =3 \] Next Question / Full Exercise

Find a − 1/a Question: If \[ a^2+\frac{1}{a^2}=102 \] find: \[ a-\frac{1}{a} \] Solution: Using identity: \[ \left(a-\frac{1}{a}\right)^2 = a^2+\frac{1}{a^2}-2 \] Substituting the given value: \[ \left(a-\frac{1}{a}\right)^2 = 102-2 \] \[ \left(a-\frac{1}{a}\right)^2 = 100 \] \[ a-\frac{1}{a} = \sqrt{100} \] \[ =10 \] Next Question / Full Exercise