Educational

Factorize x⁴ + x² + 25 Factorize x⁴ + x² + 25 Question: Factorize: \[ x^4+x^2+25 \] Solution We write \[ x^4+x^2+25 = x^4+10x^2+25-9x^2 \] \[ = (x^2+5)^2-(3x)^2 \] Using the identity \[ a^2-b^2=(a-b)(a+b) \] Therefore, \[ x^4+x^2+25 = (x^2+5-3x)(x^2+5+3x) \] \[ = (x^2-3x+5)(x^2+3x+5) \] Therefore, \[ \boxed{(x^2-3x+5)(x^2+3x+5)} \] Next Question / Full Exercise

Write the Value of 30³ + 20³ − 50³ Write the Value of 30³ + 20³ − 50³ Question: Write the value of \(30^3 + 20^3 – 50^3\). Solution Let \[ a=30,\quad b=20,\quad c=-50 \] Then, \[ a+b+c=30+20-50 \] \[ a+b+c=0 \] We know that if \(a+b+c=0\), then \[ a^3+b^3+c^3=3abc \] Therefore, \[ 30^3+20^3-50^3 =

Write the Value of (1/2)³ + (1/3)³ − (5/6)³ Write the Value of (1/2)³ + (1/3)³ − (5/6)³ Question: Write the value of \[ \left(\frac{1}{2}\right)^3 + \left(\frac{1}{3}\right)^3 – \left(\frac{5}{6}\right)^3 \] Solution Let \[ a=\frac{1}{2},\quad b=\frac{1}{3},\quad c=-\frac{5}{6} \] Then, \[ a+b+c= \frac{1}{2} + \frac{1}{3} – \frac{5}{6} \] Taking LCM = 6, \[ a+b+c= \frac{3+2-5}{6} \] \[

Write the Value of 48³ − 30³ − 18³ Write the Value of 48³ − 30³ − 18³ Question: Write the value of \(48^3 – 30^3 – 18^3\). Solution Let \[ a=48,\quad b=-30,\quad c=-18 \] Then, \[ a+b+c=48-30-18 \] \[ a+b+c=0 \] We know that if \(a+b+c=0\), then \[ a^3+b^3+c^3=3abc \] Therefore, \[ 48^3-30^3-18^3 =3(48)(-30)(-18)

Write the Value of 25³ − 75³ + 50³ Write the Value of 25³ − 75³ + 50³ Question: Write the value of \(25^3 – 75^3 + 50^3\). Solution We know that \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Let \[ a=25,\quad b=-75,\quad c=50 \] Then, \[ a+b+c=25-75+50 \] \[ a+b+c=0 \] Therefore, \[ a^3+b^3+c^3=3abc \] Substituting the values:

If a² + b² + c² = 250 and ab + bc + ca = 3, Find a + b + c If a² + b² + c² = 250 and ab + bc + ca = 3, Find a + b + c Question: If \(a^2+b^2+c^2=250\) and \(ab+bc+ca=3\), find \(a+b+c\). Solution We know that

If a + b + c = 9 and ab + bc + ca = 40, Find a² + b² + c² If a + b + c = 9 and ab + bc + ca = 40, Find a² + b² + c² Question: If \(a+b+c=9\) and \(ab+bc+ca=40\), find \(a^2+b^2+c^2\). Solution We know that

If a² + b² + c² = 20 and a + b + c = 0, Find ab + bc + ca If a² + b² + c² = 20 and a + b + c = 0, Find ab + bc + ca Question: If \(a^2+b^2+c^2=20\) and \(a+b+c=0\), find \(ab+bc+ca\). Solution We know that

If a + b + c = 0, Then Find the Value of a³ + b³ + c³ If a + b + c = 0, Then Find the Value of a³ + b³ + c³ Question: If \(a + b + c = 0\), then write the value of \(a^3 + b^3 + c^3\).

Prove that a²/bc + b²/ca + c²/ab = 3 If \(a, b, c\) are all non-zero and \(a+b+c=0\), prove that \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3 \] \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} \] \[ = \frac{a^3+b^3+c^3}{abc} \] Since, \[ a+b+c=0 \] Using the identity \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Therefore, \[ a^3+b^3+c^3-3abc=0 \] \[ a^3+b^3+c^3=3abc \] Hence, \[ \frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} \]