Educational

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

Find 4x² + 4/x² Question: If \[ x-\frac{1}{x}=\frac12 \] find: \[ 4x^2+\frac{4}{x^2} \] Solution: Using identity: \[ \left(x-\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}-2 \] Substituting the given value: \[ \left(\frac12\right)^2 = x^2+\frac{1}{x^2}-2 \] \[ \frac14 = x^2+\frac{1}{x^2}-2 \] \[ x^2+\frac{1}{x^2} = 2+\frac14 \] \[ = \frac94 \] Multiplying both sides by 4: \[ 4\left(x^2+\frac{1}{x^2}\right) = 4\times\frac94 \] \[

Find x⁶ + 1/x⁶ Question: If \[ x+\frac{1}{x}=3 \] find: \[ x^6+\frac{1}{x^6} \] Solution: First find \[ x^2+\frac{1}{x^2} \] Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] \[ 3^2 = x^2+\frac{1}{x^2}+2 \] \[ 9 = x^2+\frac{1}{x^2}+2 \] \[ x^2+\frac{1}{x^2} = 7 \] Now find \[ x^3+\frac{1}{x^3} \] Using identity: \[ \left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3} + 3\left(x+\frac{1}{x}\right) \]

Find a² + b² Question: If \[ a+b=7 \] and \[ ab=12 \] find: \[ a^2+b^2 \] Solution: Using identity: \[ (a+b)^2 = a^2+b^2+2ab \] Substituting the given values: \[ (7)^2 = a^2+b^2+2(12) \] \[ 49 = a^2+b^2+24 \] \[ a^2+b^2 = 49-24 \] \[ =25 \] Next Question / Full Exercise

Find a² + b² Question: If \[ a-b=5 \] and \[ ab=12 \] find: \[ a^2+b^2 \] Solution: Using identity: \[ (a-b)^2 = a^2+b^2-2ab \] Substituting the given values: \[ (5)^2 = a^2+b^2-2(12) \] \[ 25 = a^2+b^2-24 \] \[ a^2+b^2 = 25+24 \] \[ =49 \] Next Question / Full Exercise

Find a³ + b³ + c³ − 3abc Question: If \[ a+b+c=9 \] and \[ a^2+b^2+c^2=35 \] find: \[ a^3+b^3+c^3-3abc \] Solution: Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] First find \[ ab+bc+ca \] Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Substituting the given values: \[ 9^2 = 35+2(ab+bc+ca) \] \[ 81 = 35+2(ab+bc+ca) \]

Find x² + 1/x² Question: If \[ x+\frac{1}{x}=3 \] find: \[ x^2+\frac{1}{x^2} \] Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] Substituting the given value: \[ (3)^2 = x^2+\frac{1}{x^2}+2 \] \[ 9 = x^2+\frac{1}{x^2}+2 \] \[ x^2+\frac{1}{x^2} = 9-2 \] \[ =7 \] Next Question / Full Exercise

Find a³ + b³ + c³ − 3abc Question: If \[ a+b+c=9 \] and \[ a^2+b^2+c^2=26 \] find: \[ a^3+b^3+c^3-3abc \] Solution: Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] First find \[ ab+bc+ca \] Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Substituting the given values: \[ 9^2 = 26+2(ab+bc+ca) \] \[ 81 = 26+2(ab+bc+ca) \]