4 Marks Questions

Question 14 Marks

Prove that $(a + b + c)^3 - a^3 - b^3 - c^3 = 3(a + b)(b + c)(c + a).$

Answer

$(a + b + c)^3 = [(a + b + c)]^3 = (a + b)^3 + c^3 + 3(a + b)c(a + b + c)$
$\Rightarrow (a + b + c)^3 = a^3 + b^3 + 3ab(a + b) + c^3 + 3(a + b)c(a + b + c)$
$\Rightarrow (a + b + c)^3 - a^3 + b^3 - c^3 = 3ab(a + b) + 3(a + b)c(a + b + c)$
$\Rightarrow (a + b + c)^3 - a^3 + b^3 - c^3 = 3(a + b)[ab + ca + cb + c^2]$
$\Rightarrow (a + b + c)^3 - a^3 + b^3 - c^3 = 3(a + b)[a(b + c) + c(b + c)]$
$\Rightarrow (a + b + c)^3 - a^3 + b^3 - c^3 = 3(a + b)(b + c)(a +c)$

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Question 24 Marks

Factorise $: a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3$

Answer

We have:
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3 =$
$[a(b - c)]^3 + [a(b - c)]^3 + [b(c - a)]^3 + [c(a - b)]^3$
Put,
$a(b - c) = x, b(c - a) = y, c(a - b) = z$
Here,
$x + y + z = a(b - c) + b(c - a) + c(a - b)$
$= ab - ac + bc - ab - ab + ac - bc$
Thus,
We have:
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3 = x^3 + y^3 + z^3$
$= 3xyz [$When $x + y + z = 0, x^3 + y^3 + z^3 = 3xyz]$
$= 3a(b - c)b(c - a)c(a -b)$
$= 3abc(a - b)(b - c)(c - a)$

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Question 34 Marks

Evaluate $: (-12)^3 + 7^3 + 5^3$

Answer

$(-12)^3 + 7^3 + 5^3$
We know:
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
$x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$
Here, $x = (-12), y = 7, z = 5$
$(-12)^3 + 7^3 + 5^3$
$= (-12 + 7 + 5)[(-12)^2 + 7^2 + 5^2 - 7(-12) - 35 + 60] + 3(-12) \times 35$
$= 0 - 1260$
$= -1260$

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Question 44 Marks

Evaluate $: (28)^3 + (-15)^3 + (-13)^3$

Answer

$(28)^3 + (-15)^3 + (-13)^3$
We know:
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
$x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$
Here, $x = (-28), y = -15, z = -13$
$(28)^3 + (-15)^3 + (-13)^3$
$= (28 - 15 - 13)[(28)^2 + (-15)^2 + (-13)^2 - 28(-15) - (-15)(-13) - 28(-13)] + 3 \times 28(-15)(-13)$
$= 0 + 16380$
$= 16380$

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MATHS 4 Marks Questions

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